CBSE Class 10 Answered
& Resistances of three resistors are given as R, = 10 2, R= 20 2 and R, = 30 12. Calculate the effective resistance when they are connected in series. Also calculate the current flowing when the combination is connected to a 6 V battery.
Asked by agrimabhatt | 01 Nov, 2022, 12:26: AM
Expert Answer
Resistances are assumed as 102 Ω , 202 Ω and 3012 Ω .
Equivalenet resistance of series combination of above resistors = ( 102 + 202 +3012 ) Ω = 3316 Ω
Current drawn for battery = voltage / resistance = 6 V /3316 Ω = 1.809 × 10-3 A = 1.809 mA
Answered by Thiyagarajan K | 02 Nov, 2022, 05:26: PM
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