Question

A resistor R is connected in series before a parallel combination of two resistors of value
12 Ω and 8 Ω. Current supplied to the circuit is 4 A with supply voltage of 24 V. Find the
value of R.

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Answer 1

GIVEN :-

• In a circuit , there is a parallel combination of two resistors 12Ω and 8Ω .
• A resistor R is connected in series combination with them.
• Current supplied is 4A.
• Voltage supply is 24V.

TO FIND :-

• Unknown resistor R.

TO KNOW :-

★ For Parallel Combination ,

R(p) = [ R₁.R₂/R₁+R₂ ]

★ For Series Combination ,

R(s) = R₁ + R₂

★ By Ohm's law ,

V = IR(eq)

Here ,

• R → Resistance
• V → Voltage
• I → Current

SOLUTION :-

Resistors of 12Ω and 8Ω are in parallel combination.

So, R(p) = [ 12×8 / 12+8 ]

R(p) = 96/20

R(p) = 4.8Ω

Also , these two resistors are in series with unknown resistor R.

R(s) = R + R(p)

R(s) = [ R + 4.8 ]Ω

So, total resistance R(eq) is [ R + 4.8 ]Ω.

Now by Ohm's Law ,

V = IR

We have ,

• V = 24V
• I = 4A
• R(eq) = [R + 4.8] Ω

Putting values we get...

→ 24 = 4 [ R + 4.8 ]

→ 24/4 = R + 4.8

→ 6 = R + 4.8

→ R = 6 - 4.8

→ R = 1.2Ω

Hence , unknown resistance R is 1.2Ω .