Question

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?

Answer 1

(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