Question

To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between\(500\) and\(1650{\rm{ }}kHz\). This is accomplished with a fixed\(1.00{\rm{ }}\mu H\)inductor connected to a variable capacitor. What range of capacitance is needed?

Short answer

The range of capacitance needed is \(9.3nFn\) to \(101.3nF\).

Step-by-step solution

Definition of circuit

An electrical circuit is a closed channel that allows electricity to go from one place to another. It might also have additional electrical components including resistors, capacitors, and transistors.

Given

To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between \(500\) and \(1650{\rm{ }}kHz\). This is accomplished with a fixed \(1.00{\rm{ }}\mu H\)inductor connected to a variable capacitor.

Formula used

Using the formula for an LC circuit, the resonant frequency

\(\begin{aligned} {f_0} &= \frac{1}{{2\pi \sqrt {lC} }}\\ \Rightarrow C &= \frac{1}{{4{\pi ^2}l{f^2}_b}}\end{aligned}\)

Here, \(L\) is the self-inductance of the inductor and \(C\) is the capacitance.

Finding the range of capacitance

For min, we have\(L = 1.00 \times {10^{ - 6}}H\) and\({f_0} = 1.650 \times {10^6}\;Hz\)

Substituting the given values in above equation, we get

\(\begin{aligned} {C_{\min {\rm{ }}}} &= \frac{1}{{4 \times {{(3.14)}^2}\sqrt {\left( {1.00 \times {{10}^{ - 6}}H} \right)\left( {1.650 \times {{10}^5}\;Hz} \right)} }}\\ \Rightarrow {C_{\min }} &= 9.304 \times {10^{ - 9}}\;\\F &= 9.3nFn\end{aligned}\)

For max, we have\(L = 1.00 \times {10^{ - 6}}H\) and\({f_0} = 1.650 \times {10^6}\;Hz\)

Substituting the given values in above equation, we get

\(\begin{aligned} {C_{\max }} &= \frac{1}{{4 \times {{(3.14)}^2}\sqrt {\left( {1.00 \times {{10}^{ - 6}}H} \right){{\left( {5.001 \times {{10}^3}Hx} \right)}^2}} }}\\ \Rightarrow {C_{\max }} &= 1.013 \times {10^{ - 7}}\;\\F &= 101.3nF\end{aligned}\)

Therefore, the range of capacitance is \(9.3nFn\) to \(101.3nF\).