I. A wire of resistance 5.6 ? is connected to a battery whose emf ? is 2.0 V and

Question

I. A wire of resistance 5.6 ? is connected to a battery whose emf ? is 2.0 V and whose internal resistance is 0.44 ?. In 2.0 min, how much energy is (a) transferred from chemical to electrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

II. The current in a single-loop circuit with one resistance R is 7.3 A. When an additional resistance of 2.2 ? is inserted in series with R, the current drops to 6.1 A. What is R?

III. When resistors 1 and 2 are connected in series, the equivalent resistance is 17.1 ?. When they are connected in parallel, the equivalent resistance is 2.85 ?. What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

Explanation / Answer

First Question -

(a) Total resistance, R = 5.6 + 0.44 = 6.04 ohm

emf E = 2.0 V

So, current in the circuit, I = E / R = 2.0 / 6.04 = 0.331 A

time, t = 2 min = 2 x 60 = 120 s

Total energy transferred from chemical to electrical form = I^2*R*t = 0.331^2*6.04*120 = 79.5 J

(b) Dissipated as thermal energy in wire = 0.331^2*5.6*120 = 73.6 J

(c) Dissipated as thermal energy in the battery = 0.331^2*0.44*120 = 5.8 J

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