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:

Given:
Potential difference, V=16 V
Power consumed, P=4 W

Step 1: Equivalent Resistance of Network

• $$Left\ block:\ 4R∥4R=\ \frac{\ 4R\cdot4R}{4R\ +\ 4R}​=2R$$
• $$Right\ block:\ 6R∥12R=\ \frac{\ 6R\cdot12R}{6R\ +\ 12R}=4R$$

Total series resistance:

$$R_{eq}=2R+R+4R+R=8R$$

Step 2: Use Power Formula

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

$$4=\frac{16^2}{8R}$$

Step 3: Solve for R

$$4=\ \frac{\ 256}{8R}=\ \frac{\ 32}{R}⇒R=8Ω$$

Final Answer: 8 Ω