An $A C$ voltage is applied across a series combination of $L$ and $R$ . If the voltage drop across the resistor and inductor are $20 V$ and $15 V$ respectively, then applied peak voltage is :

25 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

35 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 \sqrt{2} V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 \sqrt{7} V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $25 \sqrt{2} V$
$V = \sqrt{V_{R}^{2} + V_{L}^{2}}$
$= \sqrt{(20)^{2} + (15)^{2}} = 25 V$
But this is the rms value.
$∴$ Peak value $= \sqrt{2} V_{r m s} = 25 \sqrt{2} V$

Suggest Corrections

Similar questions

Q. Voltage across inductor, capacitor and resistor in a series LCR ac circuit are 20V, 60 V and 70 V respectively peak value of the applied voltage is:-

Q. In a series LCR circuit the voltage across an inductor, capacitor and resistor are 20V, 20V and 40V respectively. The phase difference between the applied voltage and the current in the circuit is

Q. In a series LR circuit, the voltage drop across the inductor is 8 volt and across the resistor is 6 volt. The voltage applied and the power factor of the circuit respectively are

Q. In a series resonant circuit, the AC voltage across resistance R, inductor L and capacitor C are 5 V, 10 V and 10 V respectively. The AC voltage applied to the circuit will be

In a series resonant circuit the ac voltages across R, inductance L and capacitance C are 5V, 10 V and 10V respectively. The ac voltage applied to the circuit will be:

Join BYJU'S Learning Program

An AC voltage is applied across a series combination of L and R. If the voltage drop across the resistor and inductor are 20V and 15V respectively, then applied peak voltage is :

The correct option is B 25√2VV=√V2R+V2L=√(20)2+(15)2=25VBut this is the rms value.∴ Peak value =√2Vrms=25√2V

An $A C$ voltage is applied across a series combination of $L$ and $R$ . If the voltage drop across the resistor and inductor are $20 V$ and $15 V$ respectively, then applied peak voltage is :

The correct option is B $25 \sqrt{2} V$
$V = \sqrt{V_{R}^{2} + V_{L}^{2}}$
$= \sqrt{(20)^{2} + (15)^{2}} = 25 V$
But this is the rms value.
$∴$ Peak value $= \sqrt{2} V_{r m s} = 25 \sqrt{2} V$