- Charges forced through ε 1 will lose, rather than gain, 4 V of potential, so
the overall emf of this circuit is ε 2 − ε 1 = 8 V.
- Since the total resistance is 3 Ω + 1 Ω = 4 Ω, the current will be
.
- Finally, energy will be dissipated in these resistors at a rate of I^2 R 1 +
I^2 R 2 = (2 A)^2 (3 Ω) + (2 A)^2 (1 Ω) = 16 W. ε 2 will provide energy at a rate of
P 2 = IV 2 = (2 A)(12 V) = 24 W, while ε 1 will absorb at a rate of P 1 = IV 1 = (2
A)(4 V) = 8 W. Once again, energy is conserved; the power delivered (24 W)
equals the power taken
(8 W + 16 W = 24 W).
- All real batteries contain internal resistance, r. Determine the current in
the following circuit when the switch S is closed: