Two batteries one of emf 3 V, internal resistance 1 ohm and the other of emf 15 V, internal resistance 2 ohm are connected in series with a resistance R as shown. If the potential difference between a and b is zero, the resistance of R in ohms is
A
5
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B
7
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C
3
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D
1
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Solution
The correct option is C 3
From circuit analysis, we get
Current flow in circuit, i=EnetR+r i=18R+3
This current moves in the circuit from point b to a.
Applying KVL, Vb=−18R+3(1)+3+Va Vb−Va=0=−18+3R+9 [given:Vb−Va=0] ⇒3R=9 R=3Ω