The figure below shows a circuit with a 12.0-V battery connected to four resistors. How much power is delivered to each resistor? (Consider the following values: R1 = 1.35 Ω, R2 = 2.15 Ω, R3 = 3.95 Ω, R4 = 4.80 Ω.)

www.webassign.net/katzpse1/29-p-046-alt.png

R3 and R4 combine to produce R such that

1/R = 1/3.95 + 1/4.80 = 2.17
So, the current through R1 and R2 is 12/2.17 = 5.53A

See what you can do with that. Either find the voltage across R3 and R4, or split the current between them.

To calculate the power delivered to each resistor in this circuit, we can use the formula P = I^2 * R, where P is power, I is current, and R is the resistance.

To find the current flowing through the circuit, we need to use Ohm's law: V = I * R, where V is voltage, I is current, and R is resistance.

In this case, the voltage of the battery is 12.0 V. We need to find the total resistance of the circuit. Since the resistors are in series, the total resistance (R_total) is equal to the sum of the individual resistances: R_total = R1 + R2 + R3 + R4.

Plugging in the given values: R_total = 1.35 Ω + 2.15 Ω + 3.95 Ω + 4.80 Ω.

Now, we can use Ohm's Law to find the current (I): I = V / R_total.

Plugging in the values: I = 12.0 V / (1.35 Ω + 2.15 Ω + 3.95 Ω + 4.80 Ω).

Once we have the value of the current (I), we can calculate the power delivered to each resistor using the formula P = I^2 * R.

For example, for resistor R1: P1 = I^2 * R1.

Repeat this calculation for each resistor (R2, R3, R4) using the same formula.

Finally, you will have the power delivered to each resistor in the circuit.