Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential difference of 4.0 V. When the two resistors are connected in parallel across the 6.0 V battery, the current in B is found to be 2.0 A. Find the resistances of A and B.

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

raphealnwobi raphealnwobi · Answer 2 · 2022-03-15T13:11:20+01:00

The resistance of A is 6Ω while for B is 3Ω.

Definition of potential difference

Potential difference is the amount of work done in moving a unit charge from one point to another.

When the two resistors are connected in parallel across the 6.0 V battery, the current in B is found to be 2.0 A, hence:

2 * Rb = 6 V

Rb = 3Ω

For the series connection:

I * 3 = (6 - 4)

I = 2/3 A

Hence:

(2/3) * Ra = 4

Ra = 6Ω

The resistance of A is 6Ω while for B is 3Ω.

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