Question
a cell of emf 10v and internal resistance 2ohms is connected in series with an ammeter resistance 0.5ohms and a load of resistance 7ohms. Calculate (A) the potential difference across its terminal (B)the voltage drop
Answers
Voltage= Current X Resistance(ohms law)
V-the battery voltage
R-resistance in a circuit
V=IR
P=IV
P=V2/R, where 2 is an exponent
P=I2R, where 2 is an exponent
P=Power
q=CV
q=charge
C=Capacitance
V=voltage
(ohms law for capacitors)
V-the battery voltage
R-resistance in a circuit
V=IR
P=IV
P=V2/R, where 2 is an exponent
P=I2R, where 2 is an exponent
P=Power
q=CV
q=charge
C=Capacitance
V=voltage
(ohms law for capacitors)
Nt now.leta
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