Question

(a) Find the current in each resistor. Apply Ohm's law, solve for the current I delivered by the battery to find the current in each resistor. I_1 = frac{Delta V}{R_1} = frac{18 V}{3.0 Omega} = A I_2 = frac{Delta V}{R_2} = frac{18 V}{6.0 Omega} = A I_3 = frac{Delta V}{R_3} = frac{18 V}{9.0 Omega} = A (b) Calculate the power delivered to each resistor and the total power. Apply mathcal{P} = I^2R to each resistor, substituting the results from part (a). 3 Omega: mathcal{P}_1 = I_1^2R_1 = (I_1)^2(3.0 Omega) = W 6 Omega: mathcal{P}_2 = I_2^2R_2 = (I_2)^2(6.0 Omega) = W 9 Omega: mathcal{P}_3 = I_3^2R_3 = (I_3)^2(9.0 Omega) = W Sum to get the total power. mathcal{P}_{tot} = mathcal{P}_1 + mathcal{P}_2 + mathcal{P}_3 = imes 10^2 W (c) Find the equivalent resistance of the circuit. Apply the reciprocal-sum rule, Equation 18.6. frac{1}{R_{eq}} = frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} frac{1}{R_{eq}} = frac{1}{3.0 Omega} + frac{1}{6.0 Omega} + frac{1}{9.0 Omega} = frac{11}{18 Omega} R_{eq} = frac{18 Omega}{11} = Omega

          (a) Find the current in each resistor.
Apply Ohm's law, solve for the current I delivered by the battery to find the current in each resistor.
I_1 = frac{Delta V}{R_1} = frac{18 V}{3.0 Omega} = A
I_2 = frac{Delta V}{R_2} = frac{18 V}{6.0 Omega} = A
I_3 = frac{Delta V}{R_3} = frac{18 V}{9.0 Omega} = A
(b) Calculate the power delivered to each resistor and the total power.
Apply mathcal{P} = I^2R to each resistor, substituting the results from part (a).
3 Omega: mathcal{P}_1 = I_1^2R_1 = (I_1)^2(3.0 Omega) = W
6 Omega: mathcal{P}_2 = I_2^2R_2 = (I_2)^2(6.0 Omega) = W
9 Omega: mathcal{P}_3 = I_3^2R_3 = (I_3)^2(9.0 Omega) = W
Sum to get the total power.
mathcal{P}_{tot} = mathcal{P}_1 + mathcal{P}_2 + mathcal{P}_3 = 	imes 10^2 W
(c) Find the equivalent resistance of the circuit.
Apply the reciprocal-sum rule, Equation 18.6.
frac{1}{R_{eq}} = frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3}
frac{1}{R_{eq}} = frac{1}{3.0 Omega} + frac{1}{6.0 Omega} + frac{1}{9.0 Omega} = frac{11}{18 Omega}
R_{eq} = frac{18 Omega}{11} = Omega

$(a) Find the current in each resistor. Apply Ohm's law, solve for the current I delivered by the battery to find the current in each resistor. I1 = fracDelta VR1 = frac18 V3.0 Omega = A I2 = fracDelta VR2 = frac18 V6.0 Omega = A I3 = fracDelta VR3 = frac18 V9.0 Omega = A (b) Calculate the power delivered to each resistor and the total power. Apply mathcalP = I^2R to each resistor, substituting the results from part (a). 3 Omega: mathcalP1 = I1^2R1 = (I1)^2(3.0 Omega) = W 6 Omega: mathcalP2 = I2^2R2 = (I2)^2(6.0 Omega) = W 9 Omega: mathcalP3 = I3^2R3 = (I3)^2(9.0 Omega) = W Sum to get the total power. mathcalPtot = mathcalP1 + mathcalP2 + mathcalP3 = imes 10^2 W (c) Find the equivalent resistance of the circuit. Apply the reciprocal-sum rule, Equation 18.6. frac1Req = frac1R1 + frac1R2 + frac1R3 frac1Req = frac13.0 Omega + frac16.0 Omega + frac19.0 Omega = frac1118 Omega Req = frac18 Omega11 = Omega$

Added by Jacob D.

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