Q: In the circuit shown, the current in the 1\$\Omega\$ resistor is :

\$\textbf{Step 1 : Applying KVL in loop 1} \$\$9-2i-1(i-i_1)-3i=0\$ \$\Rightarrow 6i-i_1=9 \$ \$....(1)\$\$\textbf{Step 2 : Applying KVL in loop 2} \$\$6-3i_1+1(i-i_1)=0\$ \$\Rightarrow -i+4i_1=6 \$ \$...(2)\$\$\textbf{Step 3 : Solving equations (1) and (2)} \$Eq \$(1) \times 4 +\$ Eq\$(2)\$\$\Rightarrow 23i=42\$\$\Rightarrow i=1.826 A\$ Put the values of \$i\$ in equation \$(1)\$\$\Rightarrow 6\times 1.826 -i_1= 9\$\$\Rightarrow i_1=1.956 A\$Current through 1\$ \Omega\$ resistor \$=i-i_1=-0.13 A\$Therefore, current through 1\$ \Omega\$ resistor is \$0.13 A\$ from Q to PHence, Option (C) is correct.

Q: Find the current flowing through the following electric circuit:

Since, resistors \$1 \Omega\$ and \$3 \Omega\$ are in series,\$\therefore { R }_{ eq1 }= 1+3=4\Omega\$Now, resistors \$4 \Omega\$ and \$6\Omega\$ are in parallel,\$\therefore { R }_{ eq2 }= 4 \Omega \parallel 6 \Omega= \cfrac{ 4 \times 6 }{ 4 + 6 }= \cfrac{ 24 }{ 10 }= 2.4 \Omega\$Now, \$3.6 \Omega, 2.4 \Omega\$ and \$3 \Omega\$ are again series,\$\therefore { R }_{ eq }= 3.6+2.4+3 = 9 \Omega\$Now, the current flowing the electric current \$I\$\$I= \cfrac{ V }{ { R }_{ eq }}\$\$\Rightarrow I= \cfrac{4.9}{9}= 0.5 Amp\$\$\Rightarrow I= 0.5 Amp\$

Q: Find out the value of current through \$2\Omega\$ resistance for the given circuit.

Hint: Apply Kirchhoff’s Current Law (KCL). Correct Answer: Option A Explanation of Correct Answer: Step 1: Assume current through each branch. In the figure, the left loop is considered. Current \$I_1\$ is passing through the battery and through the \$5 \Omega \$ resistor and current \$I_2\$ is passing through \$2 \Omega \$ resistor. Step 2: Apply Kirchhoff’s Current Law (KCL) at junction A. By applying KCL, \$I_1=I_2+I_1\$ \$\Rightarrow I_2=0\$ Step 3: Apply Kirchhoff’s Current Law (KCL) at junction B.In the figure, the right loop is considered. Current \$I_1\$ is passing through the battery and through the \$10 \Omega \$ resistor and current \$I_2\$ is passing through \$2 \Omega \$ resistor. By applying KCL, \$I_1=I_2+I_1\$ \$\Rightarrow I_2=0\$ So, the current through \$2 \Omega \$ resistance is zero.

Q: In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of resistor R will be

Applying Kirchhoff's law for loop BCDFAEB: \$500 i=12-2 \Rightarrow i=10/500=1/50\$when galvanometer shows zero deflection then no current flow through it.now for loop BCDEB: \$500i+Ri=12 \Rightarrow i=\frac{12}{R+500}\$or \$1/50=12/(R+500)\$or \$R+500=600 \Rightarrow R=100 \Omega\$

Q: In the circuit given below, the current flowing across \$5\$ ohm resistor is \$1\$ amp. Find the current flowing through the other two resistor.

In the given figure, \$5\,\Omega ,\,4\,\Omega \,\,and\,10\,\Omega \$ resistors are in parallel. In parallel combination, voltage across each resistor is same. It is given that, \$ R=5\,\Omega \$ \$ I=1\,A \$ \$ V=IR \$ \$ =5\,V \$ Current across \$4\,\Omega \$ resistor \$I=\dfrac{5}{4}\,A=1.25\,A\$ Current across \$10\,\Omega \$ resistor \$I=\dfrac{5}{10}=0.5\,A\$

Q: If power dissipated in the \$9\Omega \$ resistor in the circuit shown is 36 Watt, the potential difference across the \$2\Omega \$ resistor is :

\$P = VI = {I^2}R\$\$36 = {I^2} \times 9\$\$I = 2\,Amp\$\${I_1} = \dfrac{{6 \times {I_T}}}{{15}}\$\${I_2} = \dfrac{{9 \times {I_T}}}{{15}}\$\$\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{6}{9}} \right)\$\$\dfrac{{2 \times 9}}{6} = {I_2} = 3\,Amp\$\${I_{net}} = \left( {2 + 3} \right) = 5\$\${V_{2\Omega }} = I \times R = 5 \times 2 = 10\,Volt\$

Q: Two cells of same emf E but of different internal resistance \$r_1\$ and \$r_2\$ are connected in series with an external resistance \$R\$. The potential drop across the first cell is found to be zero. The external resistance \$R\$ is

The total current in the circuit is\$I = \dfrac{2E}{r_1 + r_2 + R}\$ ..............................(1).....\$(\$In series, \$E + E = 2E)\$Now the potential drop across first cell is\$V_1 = E - Ir_1\$As per the given condition\$V_1 = 0\$hence,\$E - Ir_1 = 0\$\$E - \left(\dfrac{2E}{r_1 + r_2 + R} \right)r_1 = 0\$ .....................from\$(1)\$\$\dfrac{2r_1E}{r_1 + r_2 + R} = E\$\$\Rightarrow r_1 + r_2 + R = 2r_1\$\$R = r_1 - r_2\$

Q: In the circuit shown in the figure, if the potential at point \$A\$ is taken to be zero, the potential at point \$B\$ is

Along the path ACDB\$V_B=V_A+1V+2\times 1V-2V=0+1+2-2=1V\$

Q: The current flowing through 3 ohm resistor is 0.8 ampere, the potential drop across the 4 ohm resistor is :

Given : \$I_1 = 0.8\$ AThus voltage drop across 3 ohm, \$V_1 = IR_1 = 0.8\times 3 = 2.4\$ voltsAs 3 ohm and 6 ohm are connected in parallel, thus voltage across them is equal.\$\therefore\$ Current through 6 ohm, \$I_2 = \dfrac{V_1}{R_2}=\dfrac{2.4}{6} =0.4\$ ACurrent through 4 ohm resistor \$I' = I_1+I_2 =0.8+0.4 = 1.2\$ A\$\therefore\$ Potential drop across 4 ohm, \$V' = I'R_3 = 1.2\times 4 = 4.8\$ volts

Question 1

In the circuit shown, the current in the 1$\Omega$ resistor is :

Accepted Answer

$\textbf{Step 1 : Applying KVL in loop 1} $$9-2i-1(i-i_1)-3i=0$  $\Rightarrow 6i-i_1=9   $                               $....(1)$$\textbf{Step 2 : Applying KVL in loop 2} $$6-3i_1+1(i-i_1)=0$ $\Rightarrow -i+4i_1=6   $                                  $...(2)$$\textbf{Step 3 : Solving equations (1) and (2)} $Eq $(1) \times 4 +$ Eq$(2)$$\Rightarrow 23i=42$$\Rightarrow i=1.826 A$ Put the values of $i$ in equation $(1)$$\Rightarrow 6\times 1.826 -i_1= 9$$\Rightarrow i_1=1.956 A$Current through 1$ \Omega$ resistor $=i-i_1=-0.13 A$Therefore, current through 1$ \Omega$ resistor is $0.13 A$ from Q to PHence, Option (C) is correct.

Question 2

Find the current flowing through the following electric circuit:

Accepted Answer

Since, resistors $1 \Omega$ and $3 \Omega$ are in series,$\therefore { R }_{ eq1 }= 1+3=4\Omega$Now, resistors $4 \Omega$ and $6\Omega$ are in parallel,$\therefore { R }_{ eq2 }= 4 \Omega \parallel 6 \Omega= \cfrac{ 4 \times 6 }{ 4 + 6 }= \cfrac{ 24 }{ 10 }= 2.4 \Omega$Now, $3.6 \Omega, 2.4 \Omega$ and $3 \Omega$ are again series,$\therefore { R }_{ eq }= 3.6+2.4+3 = 9 \Omega$Now, the current flowing the electric current $I$$I= \cfrac{ V }{ { R }_{ eq }}$$\Rightarrow I= \cfrac{4.9}{9}= 0.5 Amp$$\Rightarrow I= 0.5 Amp$

Question 3

Find out the value of current through $2\Omega$ resistance for the given circuit.

Accepted Answer

Hint: Apply&#10;Kirchhoff&#8217;s Current Law (KCL).&#10;&#10;Correct Answer: Option A&#10;&#10;Explanation of Correct Answer: &#10;&#10;Step 1: Assume current through each branch.&#10;&#10;In the figure, the left loop is considered. Current $I_1$ is passing&#10;through the battery and through the $5 \Omega $ resistor and current $I_2$ is passing through $2 \Omega $ resistor.&#10;&#10;Step 2: Apply&#10;Kirchhoff&#8217;s Current Law (KCL) at junction A.&#10;&#10;By applying&#10;KCL, $I_1=I_2+I_1$&#10;&#10;$\Rightarrow&#10;I_2=0$&#10;&#10;Step 3: Apply&#10;Kirchhoff&#8217;s Current Law (KCL) at junction B.In the figure, the right loop is considered. Current $I_1$ is passing through the battery and through the $10 \Omega $ resistor and current $I_2$ is passing through $2 \Omega $ resistor.&#10;&#10;By applying&#10;KCL, $I_1=I_2+I_1$&#10;&#10;$\Rightarrow&#10;I_2=0$&#10;&#10;So, the current through $2 \Omega $ resistance is zero.

Question 4

In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of resistor R will be

Accepted Answer

Applying Kirchhoff's law for loop BCDFAEB: $500 i=12-2 \Rightarrow i=10/500=1/50$when galvanometer shows zero deflection then no current flow through it.now for loop BCDEB: $500i+Ri=12 \Rightarrow i=\frac{12}{R+500}$or $1/50=12/(R+500)$or $R+500=600 \Rightarrow R=100 \Omega$

Question 5

In the circuit given below, the current flowing across $5$ ohm resistor is $1$ amp. Find the current flowing through the other two resistor.

Accepted Answer

In the given figure, $5\,\Omega ,\,4\,\Omega&#10;\,\,and\,10\,\Omega $ resistors are in parallel. In parallel combination,&#10;voltage across each resistor is same.&#10;&#10;It is given that,&#10;&#10;$ R=5\,\Omega  $&#10;&#10;$ I=1\,A $&#10;&#10;$ V=IR $&#10;&#10;$ =5\,V $&#10;&#10;Current across $4\,\Omega $ resistor $I=\dfrac{5}{4}\,A=1.25\,A$&#10;&#10;Current across $10\,\Omega $ resistor $I=\dfrac{5}{10}=0.5\,A$

Question 6

If power dissipated in the $9\Omega $ resistor in the circuit shown is 36 Watt, the potential difference across the $2\Omega $ resistor is :

Accepted Answer

$P = VI = {I^2}R$$36 = {I^2} \times 9$$I = 2\,Amp$${I_1} = \dfrac{{6 \times {I_T}}}{{15}}$${I_2} = \dfrac{{9 \times {I_T}}}{{15}}$$\dfrac{{{I_1}}}{{{I_2}}} = \left( {\dfrac{6}{9}} \right)$$\dfrac{{2 \times 9}}{6} = {I_2} = 3\,Amp$${I_{net}} = \left( {2 + 3} \right) = 5$${V_{2\Omega }} = I \times R = 5 \times 2 = 10\,Volt$

Question 7

Two cells of same emf E but of different internal resistance $r_{1}$ and $r_{2}$ are connected in series with an external resistance $R$ . The potential drop across the first cell is found to be zero. The external resistance $R$ is

Accepted Answer

The total current in the circuit is$I = \dfrac{2E}{r_1 + r_2 + R}$ ..............................(1).....$($In series, $E + E = 2E)$Now the potential drop across first cell is$V_1 = E - Ir_1$As per the given condition$V_1 = 0$hence,$E - Ir_1 = 0$$E - \left(\dfrac{2E}{r_1 + r_2 + R} \right)r_1 = 0$ .....................from$(1)$$\dfrac{2r_1E}{r_1 + r_2 + R} = E$$\Rightarrow r_1 + r_2 + R = 2r_1$$R = r_1 - r_2$

Question 8

In the circuit shown below,the reading of the voltmeter $V$ is

Accepted Answer

Let us assume voltmeter is ideal:-It means voltmeter has infinite resistance and current will not flow in voltmeter.${ R }_{ eq }$ in upper branch  = ${ R }_{ eq }$ in lower branch. So current will get equally divide into both branches.${ V }_{ A }-{ V }_{ B }=4\times 1\quad (1)\ { V }_{ A }-{ V }_{ D }=16\times 1 \quad (2)$Subtract 2 and 1${ V }_{ B }-{ V }_{ D }=12V$A voltmeter will read 12 V

Question 9

In the circuit shown in the figure, if the potential at point $A$ is taken to be zero, the potential at point $B$ is

Accepted Answer

Along the path ACDB$V_B=V_A+1V+2\times 1V-2V=0+1+2-2=1V$

Question 10

The current flowing through 3 ohm resistor is 0.8 ampere, the potential drop across the 4 ohm resistor is :

Accepted Answer

Given :   $I_1 = 0.8$ AThus voltage drop across 3 ohm,      $V_1 = IR_1 = 0.8\times 3 = 2.4$ voltsAs 3 ohm and 6 ohm are connected in parallel, thus voltage across them is equal.$\therefore$  Current through 6 ohm,      $I_2 = \dfrac{V_1}{R_2}=\dfrac{2.4}{6} =0.4$ ACurrent through 4 ohm resistor         $I' = I_1+I_2 =0.8+0.4 = 1.2$ A$\therefore$ Potential drop across 4 ohm,       $V' = I'R_3 = 1.2\times 4 = 4.8$ volts

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In the circuit shown, the current in the 1 $Ω$ resistor is :

Find the current flowing through the following electric circuit:

Find out the value of current through $2 Ω$ resistance for the given circuit.

In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of resistor R will be

In the circuit given below, the current flowing across $5$ ohm resistor is $1$ amp. Find the current flowing through the other two resistor.

If power dissipated in the $9 Ω$ resistor in the circuit shown is 36 Watt, the potential difference across the $2 Ω$ resistor is :

Two cells of same emf E but of different internal resistance $r_{1}$ and $r_{2}$ are connected in series with an external resistance $R$ . The potential drop across the first cell is found to be zero. The external resistance $R$ is

In the circuit shown below,the reading of the voltmeter $V$ is

In the circuit shown in the figure, if the potential at point $A$ is taken to be zero, the potential at point $B$ is

The current flowing through 3 ohm resistor is 0.8 ampere, the potential drop across the 4 ohm resistor is :

Revise with Concepts

Combination of Cells in Series and Parallel

Combination of Cells - Problem L1

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Combination of EMF

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Problem-based on Combination of cells - I

In the circuit shown, the current in the 1Ω resistor is :

Find the current flowing through the following electric circuit:

Find out the value of current through 2Ω resistance for the given circuit.

In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of resistor R will be

In the circuit given below, the current flowing across 5 ohm resistor is 1 amp. Find the current flowing through the other two resistor.

If power dissipated in the 9Ω resistor in the circuit shown is 36 Watt, the potential difference across the 2Ω resistor is :

Two cells of same emf E but of different internal resistance r1​ and r2​ are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is

In the circuit shown below,the reading of the voltmeter V is

In the circuit shown in the figure, if the potential at point A is taken to be zero, the potential at point B is

The current flowing through 3 ohm resistor is 0.8 ampere, the potential drop across the 4 ohm resistor is :

In the circuit shown, the current in the 1 $Ω$ resistor is :

Find out the value of current through $2 Ω$ resistance for the given circuit.

In the circuit given below, the current flowing across $5$ ohm resistor is $1$ amp. Find the current flowing through the other two resistor.

If power dissipated in the $9 Ω$ resistor in the circuit shown is 36 Watt, the potential difference across the $2 Ω$ resistor is :

Two cells of same emf E but of different internal resistance $r_{1}$ and $r_{2}$ are connected in series with an external resistance $R$ . The potential drop across the first cell is found to be zero. The external resistance $R$ is

In the circuit shown below,the reading of the voltmeter $V$ is

In the circuit shown in the figure, if the potential at point $A$ is taken to be zero, the potential at point $B$ is