Answer to Question #92891 in Electric Circuits for Efache

Question

1. Calculate the total energy provided e.m.f of 3.0V when it causes a steady current of 0.30A to flow for 30 minutes through an electric bulb. If the battery had an internal resistance of 2.00 ohms, calculate the heat energy dissipated in the bulb at the same time.

2. The resistance of a wire of length 66m and diameter 0.14m is 10. Calculate the conductivity of the material of the wire. (take pi=22/7)

Answer 1

$$E=Pt=UIt=(3)(0.3)(30\cdot 60)=1620\ J$$

The heat energy dissipated in the bulb:

$$Q=I^2Rt=(0.3)^2(2)(30\cdot 60)=324\ J$$

2. We have:

$$R=\frac{l}{\sigma A}$$

The conductivity of the material of the wire:

$$\sigma=\frac{l}{RA}=\frac{4l}{\pi d^2R}$$

$$\sigma=\frac{4(7)66}{22(0.14)^2(10)}=430\frac{1}{\Omega m}$$