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+2 votes
1.1k views
in Current electricity by (50 points)
reopened by

Two identical cells each of emf \( 1.5 V \) are connected in parallel across a parallel combination of two resistors each of resistance \( 20 \Omega \). A voltmeter connected in the circuit measures \( 1.2 V \). The internal resistance of each cell is : 

A \( 2.5 \Omega \) 

B \( 4 \Omega \) 

C \( 5 \Omega \) 

D \( 10 \Omega \)

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2 Answers

+1 vote
by (20 points)
edited by
voltage across  10 ohm is 1.2 v
so voltage across unknown resistor is 0.3v

x/10=0.3/1.2
x=2.5ohm

[plz comment if  answer is wrong ]
+1 vote
by (45.4k points)
edited by

V = E - i r/2

1.2 = 1.5 - i r/2

i r/2 = 1.5 - 1.2

i r/2 = 0.3

Then,

\(i=\frac{1.5}{10+\frac{r}{2}}\)

10i + i r/2 = 1.5

10i + 0.3 = 1.5

10i = 1.2

i = 0.12 A

This value put in equation (i)

\(0.12\frac{r}{2}=0.3\)

\(r=\frac{0.60}{0.12}\)

r = 5 Ω

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