Question

A $4k\Omega ,0.02W$ potetiometer is used in the circuit shown below. The minimum value of the resistance $R_s$ in order to protect the potentiometer is

• A. $2.23k\Omega$
• B. $2.71k\Omega$
• C. $3.82k\Omega$
• D. $8.92k\Omega$

Answer 1

The correct option is B $2.71k\Omega$

Current capacity (I) of the potentiometer is

$I^2=\frac{0.02}{4000}=5\times {10}^{-6}$

$I=\sqrt{5\times {10}^{-6}}=2.23\times {10}^{-3}A$

$\therefore I=\frac{V}{R_s+R_{pot.}}$

$2.23\times {10}^{-3}=\frac{15}{R_s+4000}$

$(2.23\times {10}^{-3})(R_s+4000)=15$

$2.23\times {10}^{-3}{R}_s+8.92=15$

$2.23\times {10}^{-3}{R}_s=6.08$

$R_s=\frac{6.08}{2.23\times {10}^{-3}}=2.726k\Omega$

$\simeq 2.71k\Omega$

Method 2:

Maximum allowable drop across potentiometer is

$V_{max}=\sqrt{P{R}_p}$

$\Rightarrow V_{max}=\sqrt{0.02\times 4\times 1000}=\sqrt{80}V$

$\Rightarrow I_{max}=\frac{\sqrt{80}}{4k\Omega }=\sqrt{5}mA$

Now applying KVL in loop when maximum current flows,

$15-\sqrt{5}(R_S{)}_{min}-4\sqrt{5}=0$

$\Rightarrow (R_s{)}_{min}=2.71k\Omega$