Answer:
Resistance of the ammeter is RA (480 $2)(20 92) == 19.2Q (480 £2 + 20 52) (As 480 22 and 20 Q are in parallel)
• As ammeter is in series with 40.8 22, Total resistance of the circuit is R = 40.8 Q2 + R₂ = 40.8 +19.2 = 60 Q By Ohm's law,
• Current in the circuit is V 30 V 1 =-= R 60 0212= A=0.5 A Thus the reading in the ammeter will be 0.5A
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Answers & Comments
Answer:
Answer:
Resistance of the ammeter is RA (480 $2)(20 92) == 19.2Q (480 £2 + 20 52) (As 480 22 and 20 Q are in parallel)
• As ammeter is in series with 40.8 22, Total resistance of the circuit is R = 40.8 Q2 + R₂ = 40.8 +19.2 = 60 Q By Ohm's law,
• Current in the circuit is V 30 V 1 =-= R 60 0212= A=0.5 A Thus the reading in the ammeter will be 0.5A