A battery produces a 2.3-A current when connected to an unknown resistor of resi

Question

A battery produces a 2.3-A current when connected to an unknown resistor of resistance R. When a 10- resistor is connected in series with R, the current drops to 1.5 A

Part A

Determine the emf of the battery and the resistance R. Assume the internal resistance r of the battery to be zero.

Part B

Determine the emf of the battery.

Part C

Do the fact that the internal resistance r of the battery is zero make the emf greater than or less than this actual value? Do the fact that the internal resistance  of the battery is zero make the emf greater than or less than this actual value?

Explanation / Answer

EMF = V + Ir (where r is the internal resistance)

Here as r = 0, EMF = v

A & B)

Given V = 2.3 * R ...I

and V = 1.5 (R + 10) .........II

equating I and II

2.3 * R = 1.5 * (R + 10)

1.53 * R = R + 10

0.53 * R = 10

R = 10 / 0.53 = 18.87 ohm

=> V = EMF = R * 2.3 (From equation I)

V = EMF = 18.87 * 2.3 = 43.4 V

C)

From the equation, EMF = V + Ir, v is the terminal voltage

if r = 0 (decreases) then terminal voltage increases keeping EMF constant

hence, there is not influence of internal resistance on EMF

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