A cell of emf 4.5 V is connected to a junction diode whose barrier potential is 0.7 V. If the external resistance in the circuit is 190Ω, then the current in the circuit in Ampere
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answer is 0000.02.
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Detailed Solution
Potential Drop across external resistor = V-VD = 4.5-0.7 = 3.8 VCurrent in the circuit = I=V−VDR =4.5−0.7190=3.8190 = 0.02 A
A cell of emf 4.5 V is connected to a junction diode whose barrier potential is 0.7 V. If the external resistance in the circuit is 190Ω, then the current in the circuit in Ampere