Question

An inductor is connected across an oscillating emf. The peak current through the inductor is $2.0$A. What is the peak current if:

a. The inductance $L$is doubled?

b. The peak emf $E_0$is doubled?

c. The frequency $V$is doubled?

Short answer

Part (a) Peak current is $1.0$A

Part(b) Peak current is $4.0$A

Part(c) Peak current is $1.0$A

Step-by-step solution

Part(a) Step 1 : Given information

We have been given that

The peak current through the inductor is $2.0$A

We have to find peak current if inductance $L$is doubled

Part(a) Step 2 : Simplify

As we know,

$I=\frac{V_L}{X_L}=\frac{E_0}{VL}$

Where,

E₀$=$peak emf

$V=$frequency

$L=$inductance

According to question

$I=2.0$

If inductance $L$is doubled

$I^{\text{'}}=\frac{E_0}{V\left(2L\right)}=\frac{I}{2}=\frac{2}{2}=1.0$A

Part(b) Step 1 : Given information

We have been given that

The peak current through the inductor is $2.0$A

We have to find peak current if peak emf $E_0$is doubled

Part(b) Step 2 : Simplify

As we know,

$I=\frac{E_0}{VL}=2$

Where,

$E_0=$peak emf

$V=$frequency

$L=$inductance

According to question

$I=2.0$

If peak emf $E_0$is doubled

$I^{\text{'}}=\frac{2E_0}{VL}=2I=2\times 2=4$A

Part(c) Step 1 : Given information

We have been given that

The peak current through the inductor is $2.0$A

We have to find peak current if frequency $V$is doubled

Part(c) Step 2 : Simplify

As we know,

$I=\frac{E_0}{VL}$

Where,

$E_0=$Peak emf

$V=$Frequency

$L=$ Inductance

According to question

$I=2.0$

If frequency $V$is doubled

$I^{\text{'}}=\frac{E_0}{\left(2V\right)L}=\frac{I}{2}=\frac{2}{2}=1.0$A