Question

Two batteries of emf 4 V and 8 V with internal
resistance 1 W and 2 W are connected in a circuit
with resistance of 9 W as shown in figure. The current
and potential difference between the points P and Q
are?

​

Answer 1

Since the batteries are connected in reverse polarities, the net potential applied to the circuit =8V−4V=4V

The net resistance in the circuit =R+r1+r2=9Ω+1Ω+2Ω=12Ω

Hence, the current in the circuit =4V/12Ω =1/3A

Potential difference across P and Q =IR =1/3A×9Ω =3V

Answer 2

Appropriate Question:-

Two batteries of emf 4 V and 8 V with internal resistance 1 ohm and 2 ohmare connected in a circuit with external resistance of 9 ohm as shown in figure. The current and potential difference between the points P and Q are?

Given:-

Potential differences,

• V1=8V
• V2=4V

Resistances,

• R1=1ohm
• R2=2ohm
• R3=9ohm

Potential difference:-

Here batteries are connected in reverse polarities hence net potential difference applied to the circuit,

$\boxed{\sf V {net}=V_1-V_2}$

$\\ \bf\longmapsto V_{net}=8V-4V$

$\\ \boxed{\bf\longmapsto V_{net}=4V}$

Find Equivalent resistance

$\boxed{\sf R_{eq}=R_1+R_2+R_3}$

$\\ \bf\longmapsto {R_{eq}=9+1+2}$

$\\ \bf\longmapsto R_{eq}=12\Omega$

Finding Current

Using ohms law

$\boxed{\sf \dfrac{V}{I}=R}$

$\\ \bf\longmapsto I=\dfrac{V}{R}$

$\\ \bf\longmapsto I=\dfrac{4}{12}$

$\\ \bf\longmapsto I=\dfrac{1}{3}$

$\\ \boxed{\bf\longmapsto I=3.3A}$