In Figure 19.15, the current of the circuit in the 8 ohm resistor is 0.5A. What

Question

In Figure 19.15, the current of the circuit in the 8 ohm resistor is 0.5A. What is the current in the 2ohm resistor?

Picture: http://imgur.com/PWvdd

Answer: 9.5 A

My work is below. Can someone show me an easier way to calculate? or how to remember  the process?

V=8ohm*0.5A=4v

I_16 (Current in 16 ohm)=?  -> 4v/16 ohm = 0.25 A

I_20 (Current in 20 ohm)=? -> 0.5A + 0.25 A = 0.75 A

8ohm & 16ohm are parallel

R= (8ohm*16ohm)/(16ohm+8ohm) + 20 ll (6ohm*2ohm)/(6ohm+2ohm)

Voltage for 20 ohm (v)= ? = 20 ohm*0.75 A = 15 v

Battery Volt. (V)= 4v + 15v = 19v

V=IR -> I = V/R

I=19v/2ohm=9.5 A

Explanation / Answer

Voltage across 8 ohm resistor V(8) = Ri = 8(0.5) = 4V.
Current across 16 ohm resistor i(16) = V/R = 4/16 = 0.25A.
Current through the 20 ohm resistor, i(20) = i(8) + i(16) = 0.5 + 0.25 = 0.75A.
V(20) = iR = 0.75 * 20 = 15V.
Source voltage, Vs = V(8) + V(20) = 4 + 15 = 19V.

Current through 2 ohm resistor, i(2) = V/R = 19 / 2 = 9.5A

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