In the circuit shown below, the cell has an e.m.f. of 10 V and internal resistance of 1 Ω. The other resistances are shown in the figure. The potential difference VA-VB is

Question

In the circuit shown below, the cell has an e.m.f. of 10 V and internal resistance of 1 Ω. The other resistances are shown in the figure. The potential difference VA-VB is (A) 6 V (B) 4 V (C) 2 V (D) -2 V

Tardigrade Admin · Accepted Answer

Equivalent external resistance of the given circuit, Re q=4 Ω Current given by the cell, i=(E/Re q+r)=(10/(4+1))=2 A Hence, (VA-VB)=(i/2) ×(R2-R1)=(2/2)(2-4)=-2 V

Q. In the circuit shown below, the cell has an e.m.f. of $10 V$ and internal resistance of $1 Ω$ . The other resistances are shown in the figure. The potential difference $V_{A} - V_{B}$ is

6 V

4 V

2 V

-2 V

Equivalent external resistance of the given circuit, $R_{e q} = 4 Ω$
Current given by the cell, $i = \frac{E}{R _{e q} + r} = \frac{10}{( 4 + 1 )} = 2 A$
Hence, $(V_{A} - V_{B}) = \frac{i}{2} \times (R_{2} - R_{1}) = \frac{2}{2} (2 - 4) = - 2 V$

Q. In the circuit shown below, the cell has an e.m.f. of 10V and internal resistance of 1Ω. The other resistances are shown in the figure. The potential difference VA​−VB​ is

6 V

4 V

2 V

-2 V

Equivalent external resistance of the given circuit, Req​=4Ω Current given by the cell, i=Req​+rE​=(4+1)10​=2A Hence, (VA​−VB​)=2i​×(R2​−R1​)=22​(2−4)=−2V

Q. In the circuit shown below, the cell has an e.m.f. of $10 V$ and internal resistance of $1 Ω$ . The other resistances are shown in the figure. The potential difference $V_{A} - V_{B}$ is

Equivalent external resistance of the given circuit, $R_{e q} = 4 Ω$
Current given by the cell, $i = \frac{E}{R _{e q} + r} = \frac{10}{( 4 + 1 )} = 2 A$
Hence, $(V_{A} - V_{B}) = \frac{i}{2} \times (R_{2} - R_{1}) = \frac{2}{2} (2 - 4) = - 2 V$