The circuit to the right consists of a battery (V 0 = 70.0 V) and five resistors

Question

The circuit to the right consists of a battery (V0= 70.0 V) and five resistors (R11= 111 ohms, R2= 282 ohms, R3= 363 ohms, R4= 634 ohms, and R5= 565 ohms). Find the current passing through each of the specified points.

IA = ___________ mA

IB = ___________ mA

IG = ___________ mA

The circuit to the right consists of a battery (V0= 70.0 V) and five resistors (R11= 111 ohms, R2= 282 ohms, R3= 363 ohms, R4= 634 ohms, and R5= 565 ohms). Find the current passing through each of the specified points. IA = ___________ mA IB = ___________ mA IG = ___________ mA

Explanation / Answer

five resistors (R11= 111 ohms, R2= 282 ohms, R3= 363 ohms, R4= 634 ohms, and R5= 565 ohms).

as we see that four resistor (R2,R3,R4,R5) are in parallel

so equivalent resistance 1/R=1/R2+1/R3+1/R4+1/R5

so 1/R=1/282+1/363+1/634+1/565

R=103.647 ohm

now this resistance and R1 (111) are in series so

total resistance =103.647+111=214.647 ohm

so total current = 70/214.647=0.326 A

now voltage across parallel circuit will be equal and it will be =70-0.326*111=33.8 volt

so current through point A=current in R2 =33.8/282

IA=0.11986 A=119.86 mA

current in R3=33.8/363=0.0931A=93.1 mA

so current through B = 119.86+93.1

IB=212.96 mA

similarly current through G= I4+I5=33.8/634+33.8/565

IG=0.1131 A=113.1 mA

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