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: 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

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\$

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

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

Question 10

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$V_B=V_A+1V+2\times 1V-2V=0+1+2-2=1V$

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