Multiple Choice Solutions
1. A; Let denote the resistivity of silver and let As denote the cross-sectional area of silver wire. Then
Rb
=ρbL/Ab
=(5ρs)L/4^2As
=5ρsL/16As
=5Rs/16
2. D; The equation I=V/R implies that increase V by a factor of 2 will cause I to increase by a factor of 2.
3. C; Use the equation P=V^2/R:
P=V^2/R
R=V^2/P
=(120V)^2/60W
=240 Ω
4. B; The current through the circuit is
I=ϵ/(r+R)
=40V/(5 Ω + 15 Ω)=2A
5. D; The equation P=I^2R gives
P=(0.5A)^2(100 Ω) = 25 W = 25 J/s
Therefore, in 20 s, the energy dissipated as heat is
E=Pt=(25J/s)(20s)=500J
Rb
=ρbL/Ab
=(5ρs)L/4^2As
=5ρsL/16As
=5Rs/16
2. D; The equation I=V/R implies that increase V by a factor of 2 will cause I to increase by a factor of 2.
3. C; Use the equation P=V^2/R:
P=V^2/R
R=V^2/P
=(120V)^2/60W
=240 Ω
4. B; The current through the circuit is
I=ϵ/(r+R)
=40V/(5 Ω + 15 Ω)=2A
5. D; The equation P=I^2R gives
P=(0.5A)^2(100 Ω) = 25 W = 25 J/s
Therefore, in 20 s, the energy dissipated as heat is
E=Pt=(25J/s)(20s)=500J