Question #240573

A 12 ohm resistor is connected in parallel with series combinations of resistors of 8 and 16 ohms. If the drop across the 8 ohm resistor os 48 volts, determine the total impressed emf and total current.

Expert's answer

According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is

calculated by using the equation V=IR, where I is current in amps (A) and R is the resistance in

ohms (Ω).

So the voltage drop across R 1 is V 1 =IR 1 , across R 2 is V 2 =IR 2 , and across R 3 is V 3 =IR 3 . The sum of

the voltages would equal: V=V 1 +V 2 +V 3 , based on the conservation of energy and charge. If we

substitute the values for individual voltages, we get:

V=IR1+IR2+IR3V=IR1+IR2+IR3

This implies that the total resistance in a series is equal to the sum of the individual resistances.

Therefore, for every circuit with N number of resistors connected in series:

RN (series)=R1+R2+R3+…+RN.

Given that:

With 2 ohm resistor

Drop across the 8 ohm resistor os 48 volts,

Series combinations of resistors of 8 and 16 ohm

We can substitute this to determine:

total impressed emf and total current to be:

emf is 200 - / 100 and the total current, 10 - / 20

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