Calculate the value of R1 in the circuit given in Fig. 3.73 such that the circuit will resonate.

Question

Calculate the value of [latex]R_1[/latex] in the circuit given in Fig. 3.73 such that the circuit will resonate.

Accepted Answer

We know that at resonance the impedance of the circuit will be resistive only. We will calculate the value of impedance in a complex form and equate its imaginary part to zero to determine the value of [latex]R_1[/latex] .
Here, Z= [latex]R_1[/latex] + j6 and [latex]Z_2[/latex] = 10 - j4

[latex]\begin{aligned} Z & =\frac{Z_1 Z_2}{Z_1+Z_2}=\frac{\left(R_1+j 6 ight)(10-j 4)}{\left(R_1+j 6 ight)+(10-j 4)}=\frac{\left(R_1+j 6 ight)(10-j 4)}{\left(R_1+10 ight)+j(6-4)} \ & =\frac{\left(10 R_1+24 ight)+j\left(60-4 R_1 ight)}{\left.\left[\left(R_1+10 ight)+j 2 ight]\left[R_1+10 ight)-j 2 ight]} \ & =\frac{\left.\left[\left(10 R_1+24 ight)+j\left(60-4 R_1 ight) ight]\left[R_1+10 ight)-j 2 ight]}{\left.\left[\left(R_1+10 ight)+j 2 ight]\left[R_1+10 ight)-j 2 ight]} \ & =\frac{\left[\left(10 R_1+24 ight)\left(R_1+10 ight)+2\left(60-4 R_1 ight) ight]+j\left(60-4 R_1 ight)\left(R_1+10 ight)-j 2\left(10 R_1+24 ight)}{\left(\left(R_1+10 ight)^2+2 ight)} \end{aligned}[/latex]

During resonance, the imaginary part of Z will be zero.
Therfore,

[latex]\mathrm{j}\left(60-4 \mathrm{R}_1 ight)\left(\mathrm{R}_1+10 ight)-\mathrm{j} 2\left(10 \mathrm{R}_1+24 ight)=0[/latex]

or, [latex]60 R_1-4 R_1^2+600-40 R_1-20 R_1-48=0[/latex]

or, [latex]4 \mathrm{R}_1{ }^2=552[/latex]

or, [latex]\mathrm{R}_1^2=138[/latex]

or, [latex]\mathrm{R}_1=11.74 \Omega[/latex]

Question 3.43: Calculate the value of R1 in the circuit given in Fig. 3.73 ......

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