Question

A battery with an emf of 24 volts and an internal resistance of 1 ohm is connected to an external circuit as shown above. Determine each of the following:

a. the equivalent resistance of the combination of the 4-ohm, 8-ohm, and 12-ohm resistors

b. the current in the 5-ohm resistor

c. the terminal voltage, V_AC of the battery

d. the rate at which energy is dissipated in the 12-ohm resistor 

e. the magnitude of the potential difference V_BC

f. the power delivered by the battery to the external circuit.

Answer 1

 a. The 4 Ω and 8 Ω are in series so their equivalent resistance is 12 Ω. Another 12 Ω resistor in parallel makes the equivalent resistance (12 × 12)/(12 + 12) = 6 Ω.

b. Adding the remaining resistors in series throughout the circuit gives a total circuit resistance of 12 Ω and the total current (which is also the current in the 5 Ω resistor) = ε/R = 2 A

c. V_AC = ε – Ir = 22 V

d. The current divides equally between the two branches on the right so P₁₂ = I²R = (1 A)²(12 Ω) = 12 W

e. From B to C you only have to pass through the 12 Ω resistor which gives V = (1 A)(12 Ω) = 12 V

f. P_B = V_AC²/R_external = (22 V)²/11 Ω = 44 W