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1The charge transferred is 1 C which corresponds to 1 1.6×10−19=6.2×10 18 electrons.

2 R=ρ L A. L is doubled; the radius is doubled so the area increases by 4. Hence the new resistance is R′=ρ 2 L 4 A=R 2.

3 P=V 2 R

So, 32=12 2 R⇒R=12 2 32

The new power is:

P′=V 2 R=15 2(12 2 32)=15 2 12 2×32=5 2 4 2×32=50 W.

4The total resistance increases so the current decreases.

5We are told that 1 R+1 R+1 R=1 1 which means that R = 3 Ω. So the series connection would give a total resistance of 9 Ω.

6The total resistance in the series connection is 2R. In the parallel connection it is R 2. Since resistance is reduced by 4 the power will go up by 4 (since P=V 2 R) and so the new power will be 64 W.

7The total resistance of the circuit is 6.0 Ω and so the current is I=6.0 3.0=2.0 A.

aThe power dissipated in the cell is P = r I 2 = 0.50 × 2.02 = 2.0 W.

bThe power in the external resistor is P = R I 2 = 2.5 × 2.02 = 10 W. (The total power is P = ε I = 6.0 × 2.0 = 12 W (= 10 + 2.0) as expected.)