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A 560 , a 380 , and a 325 resistor are connected in parallel across a 120V battery. What is the total resistance of the circuit? What is the circuit current? What is the current through each resistor? What is the power dissipated?

User Yehor
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1 Answer

3 votes

Answer:

(a). Total resistance of the circuit R = 133.43 ohm

(b). Current I = 0.9 A

(c). Current in each resistor
I_1 = 0.214 A ,
I_2 = 0.315 A &
I_3 = 0.369 A

(d). Power dissipated through the circuit is P = 108 W

Step-by-step explanation:

Given data


R_(1) = 560 ohm


R_2 = 380 ohm


R_3 = 325 ohm

Voltage V = 120 V

(a).Total resistance of the circuit in case of parallel circuit is given by


\frac{1} {R} = (1)/(R_1) + (1)/(R_2) + (1)/(R_3)


(1)/(R) = (1)/(560) + (1)/(380) + (1)/(325)

R = 133.43 ohm

(b). We know that from ohm's law

Voltage V = I R

120 = I × 133.43

I = 0.9 A

This is the value of current in the circuit.

(c). Current in resistor one


I_1 = (V)/(R_1)


I_1 = (120)/(560)


I_1 = 0.214 A

Current in second resistor


I_2 = (V)/(R_2)


I_2 = (120)/(380)


I_2 = 0.315 A

Current in third resistor


I_3 = (V)/(R_3)


I_3 = (120)/(325)


I_3 = 0.369 A

(d). Power dissipated through the circuit is given by

P = V I

P = 120 × 0.9

P = 108 W

User Maccard
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