Question

Please solve this briefly and explain​

Answer 1

Answer:

Hence, we get, Potential Difference across 3 Ω = 4 V

Explanation:

Given,

Given four resistors where R represents resistors.

R1 = 6 Ω

R2 = 3 Ω

R3 = 2 Ω

R4 = 4 Ω

Current, I = 2 A

Here in the figure, we see that,

R1 and R2 are connected in parallel which is connected in series with R3

Now the combination of those three resistors is connected in parallel with R4.

So,

For first parallel combination of R1 and R2,

1 / R(eq) = 1 / R1 + 1 / R2 = 1/6 + 1/3

=> 1 / R(eq) = 3/6 = 1/2

=> R(eq) = 2 Ω

Now, for series combination with R3,

=> R(eq) = 2 Ω + R3 = 2 Ω + 2 Ω = 4 Ω

Now this in parallel with R4,

1/R(eq) = 1/4 + 1/R4 = 1/4 + 1/4 = 1/2

=> R(eq) = 2 Ω

Hence, we get, the net effective equivalent resistance of the circuit = 2 Ω

Let the potential difference through the circuit be 'V'.

By ohm's Law we know that ,

Resistance = Voltage / Current

=> R = V / I

=> 2 = V / 2

=> V = 4 V

Hence, the potential difference of the circuit = 4 V

In parallel circuits, the electric potential difference across each resistor (ΔV) is the same. In a parallel circuit, the voltage drops across each of the branches is the same as the voltage gain in the battery. Thus, the voltage drop is the same across each of these resistors.

So, we get, that in parallel combination, potential difference remains same in the components.

Here,

Potential difference across R4 = 4 V

Potential difference across R1 = 4 V

Potential difference across R2 = 4 V

Hence, we get, Potential Difference across 3 Ω = 4 V