Two wires of resistance 3Ω and RΩ are connected in series to an ammeter and and a cell of
emf 6V and internal resistance 1Ω. The ammeter reads 0.5A. (i) calculate the value of R (ii)
calculate the charge passing through the 3Ω resistor in 120s. (iii) Calculate the power
dissipated in the 3Ω resistor.
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Answer:
Total R = 3+ R ohm
emf = 6V
r = 1 ohm
A = 0.5 A
emf = I(R+r)
6 = 0.5(3+R+1)
R = 8 ohm
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