An ideal coil of 10 ohm is connected in series with a resistance of 5 ohm and a battery of 5 volt.Two seconds after the connection is made , the current flowing in ampere in the circuit is
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R1 =10
R2=5
So, resistance =1/r1+1/r2
1/10+1/5=3/10
So, total resistance is 3.33 ohms
V=I*R
Hence, current in the circuit is 1.501 ampere
R2=5
So, resistance =1/r1+1/r2
1/10+1/5=3/10
So, total resistance is 3.33 ohms
V=I*R
Hence, current in the circuit is 1.501 ampere
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