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?

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

(b).  Current I = 0.9 A

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

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

Explanation:

Given data

= 560 ohm

= 380 ohm

= 325 ohm

Voltage V = 120 V

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

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

= 0.214 A

Current in  second resistor

= 0.315 A

Current in third resistor

= 0.369 A

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

P = V I

P = 120 × 0.9

P = 108 W

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq

f=ma newton

f=100x2.5=250n