In Figure ( a ), both batteries have emf ? = 1.10 V and the external resistance

Question

In Figure (a), both batteries have emf ? = 1.10 V and the external resistance R is a variable resistor. Figure (b) gives the electric potentials V between the terminals of each battery as functions of R: Curve 1 corresponds to battery 1, and curve 2 corresponds to battery 2. The horizontal scale is set byRs = 0.300 ?. What is the internal resistance of (a) battery 1 and (b) battery 2?

In Figure (a), both batteries have emf epsilon = 1. 10 V and the external resistance R is a variable resistor. Figure (b) gives the electric potentials V between the terminals of each battery as functions of R: Curve 1 corresponds to battery 1, and curve 2 corresponds to battery 2. The horizontal scale is set by Rs = 0. 300 ohm. What is the internal resistance of (a) battery 1 and (b) battery 2?

Explanation / Answer

V1 + V2 = i R

(4/5)*0.6 + 0 = i * (Rs/2)

(4/5)*0.55 + 0 = i * (0.3/2)

i = 2.9333 A

V1 = emf1 - i r1

(4/5)*0.55 = 1.10 - 2.9333 * r1

r1 = 0.225 ohms

V2 = emf2 - i r2

0 = 1.10 - 2.9333 * r2

r2 = 0.375 ohms

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