Two similar accumulators each of emf E and internal resistance r are connected as shown in the following figure. Then the potential difference between X and Y is:

Question

Two similar accumulators each of emf E and internal resistance r are connected as shown in the following figure. Then the potential difference between X and Y is: (A) 2E (B) E (C) zero (D) none of these

Tardigrade Admin · Accepted Answer

If I be the current in the circuit, then E+E =I r +I r=2 I r or I =(E/r) ∴ Vx-Vy =E-I r =E-(E/r) × r=0

Q. Two similar accumulators each of emf $E$ and internal resistance $r$ are connected as shown in the following figure. Then the potential difference between $X$ and $Y$ is:

2E

E

zero

none of these

If $I$ be the current in the circuit, then
$E + E = I r + I r = 2 I r$
or $I = \frac{E}{r}$
$∴ V_{x} - V_{y} = E - I r$
$= E - \frac{E}{r} \times r = 0$

Q. Two similar accumulators each of emf E and internal resistance r are connected as shown in the following figure. Then the potential difference between X and Y is:

2E

E

zero

none of these

If I be the current in the circuit, then E+E=Ir+Ir=2Ir or I=rE​ ∴Vx​−Vy​=E−Ir =E−rE​×r=0

Q. Two similar accumulators each of emf $E$ and internal resistance $r$ are connected as shown in the following figure. Then the potential difference between $X$ and $Y$ is:

If $I$ be the current in the circuit, then
$E + E = I r + I r = 2 I r$
or $I = \frac{E}{r}$
$∴ V_{x} - V_{y} = E - I r$
$= E - \frac{E}{r} \times r = 0$