Respuesta :
Answer:
(a). Total resistance of the circuit R = 133.43 ohm
(b). Current I = 0.9 A
(c). Current in each resistor [tex]I_1[/tex] = 0.214 A , [tex]I_2[/tex] = 0.315 A & [tex]I_3[/tex] = 0.369 A
(d). Power dissipated through the circuit is P = 108 W
Explanation:
Given data
[tex]R_{1}[/tex] = 560 ohm
[tex]R_2[/tex] = 380 ohm
[tex]R_3[/tex] = 325 ohm
Voltage V = 120 V
(a).Total resistance of the circuit in case of parallel circuit is given by
[tex]\frac{1} {R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}[/tex]
[tex]\frac{1}{R} = \frac{1}{560} + \frac{1}{380} + \frac{1}{325}[/tex]
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
[tex]I_1 = \frac{V}{R_1}[/tex]
[tex]I_1 = \frac{120}{560}[/tex]
[tex]I_1[/tex] = 0.214 A
Current in second resistor
[tex]I_2 = \frac{V}{R_2}[/tex]
[tex]I_2 = \frac{120}{380}[/tex]
[tex]I_2[/tex] = 0.315 A
Current in third resistor
[tex]I_3 = \frac{V}{R_3}[/tex]
[tex]I_3 = \frac{120}{325}[/tex]
[tex]I_3[/tex] = 0.369 A
(d). Power dissipated through the circuit is given by
P = V I
P = 120 × 0.9
P = 108 W
