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A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm in series. The potential difference between the terminals of the battery is

A
12.5 V
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B
9 V
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C
11.25 V
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D
10 V
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Solution

The correct option is C 11.25 V
Given data:
E=15 V,r=3Ω,R1=3Ω.R2=6Ω

Step 1: Draw circuit diagram and apply KVL [Refer Fig. 1]
While applying KVL in clockwise direction, increase in voltage is considered as positive and decrease in voltage as negative.

EiR1iR2ir=0 .........(1)

i=ER1+R2+r ............(2)

Step 2: Finding terminal Potential Difference of Battery [Refer Fig. 2]
While writing potential difference VAB, we go from B to A, increase in voltage is considered as positive and decrease in voltage as negative.

Vterminal=VAB=ir+E=EErR1+R2+r

=1515×33+6+3=11.25V
Vterminal=11.25 V

2101659_173870_ans_2a6325af5ef54221863e3d458d43a4ba.png

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