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.

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.

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Combination of Cells in Series and Parallel

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: 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 below,the reading of the voltmeter \$V\$ is

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

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\$

Physics

definition

Equivalent emf/resistance of cells in series

E

is the overall emf of the battery combined by

n

number cells and

E_{1}, E_{2}, E_{3}, E_{n}

are the emfs of individual cells.
Then

E = E_{1} + E_{2} + E_{3} + . . . + E_{n}

Similarly, if

r_{1}, r_{2}, r_{3}, r_{n}

are the internal resistances of individual cells, then the internal resistance of the battery will be equal to the sum of the internal resistance of the individual cells i.e.

r = r_{1} + r_{2} + r_{3} + + r_{n}

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Combination of Cells - Problem L1

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