Question

A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at $10\text{V}$. Box P contains a capacitance of $1\mu F$ in series with a resistance of $32\Omega$. Coil Q has a self inductance of 4.9 mH and a resistance of $68\Omega$ in series. The frequency is adjusted so that maximum current flows in P and Q.

The impedance of P at this frequency is

• A. $77\Omega$
• B. $36\Omega$
• C. $40\Omega$
• D. $125\Omega$

Answer 1

The correct option is A $77\Omega$

For Box P, Impedence is given by
$Z=\sqrt{R_1^2+(X_C{)}^2}$ .......(i)
Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance $\omega =\frac{1}{\sqrt{LC}}$
$\omega =\frac{1}{\sqrt{4.9\times {10}^{-3}\times {10}^{-6}}}\omega =\frac{{10}^5}{7}$
From here we can find the value of $X_C$
$X_C=\frac{1}{\omega C}=\frac{1\times 7}{{10}^5\times {10}^{-6}}=70\Omega$
Putting this and value of $R_1$ in eq(i)
$Z=\sqrt{{32}^2+{70}^2}=77\Omega$