Question

Please find the power dissipated by the resistor 2, P2 in the circuit. V_{tot} 12 V R_1 20 $\Omega$ R_2 30 $\Omega$ R_3 15 $\Omega$ Ohm's Law: V = I \times R, I = V/R, R = V/I Watt's Law: P = I \times V, P = V^2/R, P = I^2R In series circuit, R_{total} = R_1 + R_2 + R_3 + R_4 + ... + R_n In parallel circuit, $R = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + ... + \frac{1}{R_n}}$

          Please find the power dissipated by the resistor 2, P2 in the circuit.
V_{tot}
12 V
R_1
20 $\Omega$
R_2
30 $\Omega$
R_3
15 $\Omega$
Ohm's Law:
V = I \times R, I = V/R, R = V/I
Watt's Law:
P = I \times V, P = V^2/R, P = I^2R
In series circuit,
R_{total} = R_1 + R_2 + R_3 + R_4 + ... + R_n
In parallel circuit, $R = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + ... + \frac{1}{R_n}}$

Please find the power dissipated by the resistor 2, P2 in the circuit.
Vtot
12 V
R1
20 Ω
R2
30 Ω
R3
15 Ω
Ohm's Law:
V = I ×R, I = V/R, R = V/I
Watt's Law:
P = I ×V, P = V^2/R, P = I^2R
In series circuit,
Rtotal = R1 + R2 + R3 + R4 + ... + Rn
In parallel circuit, R = (1)/((1)/(R1) + (1)/(R2) + (1)/(R3) + (1)/(R4) + ... + (1)/(Rn))

Added by Jose Miguel R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prem Bijarniya

Step 1

In a parallel circuit the voltage across each resistor equals the total voltage: V_R2 = V_tot = 12 V. Show more…

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Thanks for your feedback!

Please find the power dissipated by the resistor 2, P2, in the circuit. Vtot = 12V R1 = 20Î© R2 = 30Î© R3 = 15Î© Ohm's Law: V = IR I = V/R R = V/I Watt's Law: In a series circuit, P = IV P = V^2/R P = I^2R Rtotal = R1 + R2 + R3 + ... + Rn Rtotal = R1 + R2 + R3 + R4