How is the circuit arranged in this solution?

550 ohm in series with the parallel combination of R and (50 + 500) ohm in series

550 ohm in parallel with R, all in series with 50 and 500 ohm

R, 50 ohm and 500 ohm all in parallel

550 ohm and R in series, parallel with 500 ohm

Asked 2/15
2071 Shrawan · 6 marks
2068 Shrawan · 10 marks

Unknown resistor from voltage across 500Ω

Short answer

Assuming a 12 V supply, the voltage across the parallel group is 2.75 V. Current through the 500 Ω branch is 5 mA, total current is 16.82 mA, so the unknown R = 2.75 V / 11.82 mA ≈ 233 Ω.

What is the value of the unknown resistor 'R' in figure below, if the voltage drop across 500Ω resistor is 2.5 volts? [Figure: 550 Ω in series on top; then 50 Ω in the top branch; R and 500 Ω in parallel vertical branches.]

The scanned figure of one paper shows no supply value, but the same question in the 2068 Shrawan paper shows a 12 V supply across the input terminals. I use 12 V as the assumed supply.

Circuit: 550 Ω in series with the parallel combination of R and (50 Ω + 500 Ω in series).

 o--[550]--+--[50]--+
           |        |
          [R]     [500]
           |        |
 o---------+--------+

Step 1: Current in the 50 Ω and 500 Ω branch

I₅₀₀ = 2.5/500 = 5 mA

Step 2: Voltage across the parallel group (node voltage)

V_N = 2.5 + 5 mA × 50 = 2.5 + 0.25 = 2.75 V

Step 3: Current in the 550 Ω resistor

Voltage across 550 Ω = 12 − 2.75 = 9.25 V I = 9.25/550 = 16.82 mA

Step 4: Current in R

I_R = 16.82 − 5 = 11.82 mA

Step 5: Value of R

R = 2.75/11.82 mA = 232.7 Ω

Answer: R ≈ 233 Ω (for a 12 V supply)

✓ Answer checked by Claude Opus 5.5 (high effort)

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