Question

A wire of resistance 3.0 is connected to a battery whose emf is 5.0 V and whose internal resistance is 1.0 .

(a) How much energy is transferred from chemical to electrical form in 4.0 min?
(b) How much energy appears in the wire as thermal energy in that time?

Answer 1

Answer on Question #73340 - Physics / Electric Circuits

A wire of resistance $R = 3.0 \, \Omega$ is connected to a battery whose emf is $\varepsilon = 5.0 \, \mathrm{V}$ and whose internal resistance is $1.0 \, \Omega$.

(a) How much energy is transferred from chemical to electrical form in $t = 4.0 \, \mathrm{min}$?

(b) How much energy appears in the wire as thermal energy in that time?

Solution:

(a) Total energy that is transferred from chemical to electrical form in $t = 4 \, \mathrm{min}$

$$U_{\text{total}} = \varepsilon I t$$

where $I = \frac{\varepsilon}{R + r}$ is the current through the circuit.

So

$$U_{\text{total}} = \frac{\varepsilon^2}{R + r} t = \frac{5.0^2}{3.0 + 1.0} \times 4 \times 60 = 1500 \, \mathrm{J}.$$

(b)

$$U_{\text{wire}} = I^2 R t = \frac{\varepsilon^2}{(R + r)^2} R t = \frac{5.0^2}{(3.0 + 1.0)^2} \times 3.0 \times 4 \times 60 = 1125 \, \mathrm{J}.$$

Answers:

(a) 1500 J

(b) 1125 J

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