Question

6Ω and 3 Ω resistors are connected in parallel and an 8Ω resistor is connected in series
with them. A current of 2 A passes through 8Ω resistor. Find
a) the combined resistance.
b) the potential difference across the combined resistance.
c) the current through the 3Ω resistor.

Answer 1

A 3Ω resistance and a 6Ω resistor are connected in parallel and the combination is connected in series to a battery of 5V and a 3Ω resistor. What is the potential difference across the 6Ω resistor?

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[math]\text{Equivalent resistance of the parallel resistors} = \dfrac{3\times 6}{3 + 6} = 2\,\Omega[/math]

[math]\text{Total resistance of the circuit} = 2 + 3 = 5\,\Omega[/math]

[math]\text{Current flowing in the circuit} = \dfrac{5}{5} = 1\,A[/math]

[math]\text{current passing through}\,\,6\,\Omega\,\,\text{resistor}[/math]

[math]= \dfrac{3}{3 + 6} (1) = \dfrac{1}{3}\,A[/math]

[math]\text{Voltage drop across}\,\,6\,\Omega\,\,\text{ resistor} = \dfrac{1}{3} (6) = 2 \,V[/math]

First, what's the equivalent resistance of the 3Ω and 6Ω resistors? Using the quick method, it's

(3×6)÷(3+6) or 2Ω

So your circuit's effectively 3Ω in series with 2Ω, 5Ω in total.

Current according to Ohm's Law is

I=V/R

V = 5V, R = 5Ω so I = 1A So, Ohm's Law again to calculate the voltage drop across the 3//2 combination,

V = IR, I is 1, R is 2, so the answer's 2V.

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