In the figure shown- battery 1 has emf - 6V- internal resistance - 1- Battery 2 has emf - 2V and internal resistance - 3- The wires have negligible resistance- What is the potential difference across the terminals of battery 2-

The correct option is D 5 V Using KVL, #160; 6 2=(1+3)I #160; #160;or I=1 #160; A Thus terminal potential of battery 2 is V_2= ...

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Standard XII

Physics

Simple Circuit

In the figure...

Question

In the figure shown, battery 1 has emf $=$ 6V, internal resistance $= 1 Ω$ . Battery 2 has emf $=$ 2V and internal resistance $= 3 Ω$ . The wires have negligible resistance. What is the potential difference across the terminals of battery 2?

4 V

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1.5 V

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5 V

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0.5 V

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Solution

The correct option is D $5 V$
Using KVL, $6 - 2 = (1 + 3) I$ or $I = 1 A$
Thus terminal potential of battery 2 is $V_{2} = 2 + 3 I = 2 + 3 = 5 V$

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In the figure shown, battery 1 has emf = 6V, internal resistance =1Ω. Battery 2 has emf = 2V and internal resistance =3Ω. The wires have negligible resistance. What is the potential difference across the terminals of battery 2?

The correct option is D 5VUsing KVL, 6−2=(1+3)I or I=1AThus terminal potential of battery 2 is V2=2+3I=2+3=5V

In the figure shown, battery 1 has emf $=$ 6V, internal resistance $= 1 Ω$ . Battery 2 has emf $=$ 2V and internal resistance $= 3 Ω$ . The wires have negligible resistance. What is the potential difference across the terminals of battery 2?

The correct option is D $5 V$
Using KVL, $6 - 2 = (1 + 3) I$ or $I = 1 A$
Thus terminal potential of battery 2 is $V_{2} = 2 + 3 I = 2 + 3 = 5 V$