A transformer 840 turns of wire in its primary and 660 turns of wire in its secondary. A voltage Vrms = 120 volts is applied to the primary. A current Irms = 0.16 A flows into the primary. A light bulb -- acting like an ideal resistor -- is attached to the secondary. What is the resistance of the light bulb ?
| a. |
1200 Ω |
|
| b. |
460 Ω |
|
| c. |
260 Ω |
|
| d. |
90 Ω |
|
| e. |
950 Ω |
Voltage across secondary
Vs = 660* 120 / 840
Vs = 94.29 V
Current through secondary
Is = 0.16* 840 / 660 = 0.204 A
Using ohm' s law
R = Vs/Is
R = 460 ohm
Comment in case any doubt please rate my answer....
A transformer 840 turns of wire in its primary and 660 turns of wire in its...
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