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Current Electricity question

2019 · 12 Apr · Shift 1 · Q58

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JEE Main
Physics
Current Electricity
2019 · 12 Apr · Shift 1 · Q58

Current Electricity question

2019 · 12 Apr · Shift 1 · Q58

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1

The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is: JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 243 English

JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 243 English

A
16 $\Omega$
B
1 $\Omega$
C
8 $\Omega$
D
6 $\Omega$

View written solutionFree

Correct answer: C

Find the equivalent resistance of the whole network from given power and voltage

Given:

Source voltage: $V = 16\,\text{V}$
Total power consumed: $P = 4\,\text{W}$

For the whole resistive network,

P = \frac{V^2}{R_{\text{eq}}}

So,

R_{\text{eq}} = \frac{V^2}{P} = \frac{16^2}{4} = \frac{256}{4} = 64\,\Omega

Thus, the equivalent resistance of the network must be

R_{\text{eq}} = 64\,\Omega

Use the circuit reduction

From the given circuit diagram, simplifying the resistor combination gives the total equivalent resistance in terms of $R$ as:

R_{\text{eq}} = 8R

Equate with the known equivalent resistance

Since the network equivalent resistance is $64\,\Omega$ ,

8R = 64

R = 8\,\Omega

Check the options

A: $16\,\Omega$
B: $1\,\Omega$
C: $8\,\Omega$
D: $6\,\Omega$

So the correct option is:

\boxed{8\,\Omega}

which is Option C.

Comparison with stored correct answer

Stored correct answer: C

Our derived answer is also C. Hence, they agree.

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