an emf 10v having internal resistance of 2 ohm's is connected to an external resistance of 3 ohm's the battery is first in charging mode and then in dischargeing mode 1.The current flowing through the external resistance is ? 2.the terminal potential difference during discharging mode of battery?
Answers
Answer:
The total resistance of a circuit connected to the given cell of emf 15 V is the sum of all the resistances from various sources like, resistors, ammeter, voltmeter, etc., connected in series along with the internal resistance of the cell.
In this case, the total resistance is given as 3Ω+3Ω+6Ω=12ohms.
From the ohm's law the total current can be calculated from the formula I=V/R.
That is, I=RV=12A15V=1.25A.
Hence, the current through the battery is 1.25 amperes.
The voltage output of a device is measured across its terminals and is called its terminal voltage V. Terminal voltage is given by the equation V=emf−Ir where, r is the internal resistance and I is the current flowing at the time of the measurement.
Therefore, V=15V−(1.25×3)=11.25V.
Hence, the potential difference across the terminals of the cell is 11.25V
Answer:Let the emf of the battery be E.
Internal resistance of battery r=3Ω.
Voltage across the resistor 20Ω is 10 volts.
Voltage across resistor 20Ω V=
R
RR
Now another resistor 30Ω is connected in series with battery and previous resistor.
Total resistance of the circuit R
Current flowing through the circuit I=
∴ I=
THANKS
Explanation: