Question

Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series.

a) I1 = 2 A, P1 = 96 W, I2 = 0.5 A, P2 = 48 W
b) I1 = 1 A, P1 = 48 W, I2 = 0.5 A, P2 = 24 W
c) I1 = 2 A, P1 = 96 W, I2 = 2 A, P2 = 192 W
d) I1 = 1 A, P1 = 48 W, I2 = 2 A, P2 = 96 W

Answer 1

When resistors are connected in series, the total resistance is the sum of the individual resistances. The current flowing through the circuit can be calculated using Ohm's law, and the power dissipated by each resistor can be calculated using the formula P = IV. The correct option is a) I1 = 0.4 A, P1 = 9.6 W, I2 = 0.4 A, P2 = 38.4 W.

Step-by-step explanation:

When resistors are connected in series, the total resistance is the sum of the individual resistances. In this case, the total resistance is 24.0 Ω + 96.0 Ω = 120.0 Ω. Using Ohm's law (V = IR), we can calculate the current flowing through the circuit.

The current is given by the battery voltage divided by the total resistance: I = 48.0 V / 120.0 Ω = 0.4 A. To find the power dissipated by each resistor, we can use the formula P = IV, where P is power, I is current, and V is voltage.

For the 24.0 Ω resistor, the power is P1 = 0.4 A * 24.0 Ω = 9.6 W.

For the 96.0 Ω resistor, the power is P2 = 0.4 A * 96.0 Ω = 38.4 W.

So, the correct option is a) I1 = 0.4 A, P1 = 9.6 W, I2 = 0.4 A, P2 = 38.4 W.