In a series LCR circuit, we need to find the peak voltage across the inductor given the following data:
The rms voltage across the resistor,
VR, is 30 V .
The rms voltage across the capacitor,
VC, is 90 V .
The applied voltage,
V‌applied ‌, is given by:
V‌applied ‌=50√2sin‌ωt. This means the peak value of the applied voltage is
V‌peak ‌=50√2V.
First, let's determine the rms voltage of the applied voltage. Since rms voltage is the peak voltage divided by
√2, we have:
V‌rms, applied ‌=‌=50VIn a series LCR circuit, the total rms voltage is the square root of the sum of the squares of the individual voltage drops. Since the total voltage is applied across the resistor, inductor, and capacitor, we have:
V‌total ‌=√VR2+(VL−VC)2Given:
‌VR=30V‌VC=90V‌V‌rms ‌,‌ total ‌=50VRearranging the equation, we find:
‌50=√302+(VL−90)2‌502=302+(VL−90)2‌2500=900+(VL−90)2‌2500−900=(VL−90)2‌1600=(VL−90)2 Taking the square root of both sides:
VL−90=±40This gives us two possible values for
VL :
VL=130V‌ or ‌VL=50V To find the peak voltage across the inductor, we need to multiply the rms voltage by
√2 (since peak voltage
V‌peak ‌=V‌rms ‌×√2) :
For
VL=130V:V‌peak, ‌L=130×√2=130√2VFor
VL=50V:V‌peak, L ‌=50×√2=50√2VThe correct peak voltage across the inductor that matches the options provided is
50√2V. Therefore, the answer is:
Option D:
50√2V