The battery of emf 12v and internal resistance 1.5ohm connected to an external resistance 3ohm. Calculate the current flowing in the circuit, the potential drop across the 3ohm resistance and in the internal resistance, the internal voltage of the battery, the reading of a voltmeter connected across the battery terminal in open circuit.
To calculate the current flowing in the circuit (I), we can use Ohm's Law:
I = V / R
Where V is the voltage and R is the resistance.
In this case, the total resistance in the circuit (Rtotal) is the sum of the internal resistance (Rinternal) and the external resistance (Rexternal):
Rtotal = Rinternal + Rexternal
Given that the battery's emf (E) is 12V and the internal resistance (Rinternal) is 1.5 Ohm, we have:
Rinternal = 1.5 Ohm
Rexternal = 3 Ohm
Rtotal = 1.5 Ohm + 3 Ohm = 4.5 Ohm
Now we can calculate the current (I):
I = E / Rtotal = 12V / 4.5 Ohm = 2.67 Amps
So, the current flowing in the circuit is 2.67 Amps.
To calculate the potential drop across the 3 Ohm external resistance, we use Ohm's Law again:
Voltage drop (Vdrop) = I * Rexternal = 2.67 Amps * 3 Ohm = 8.01 Volts
Therefore, the potential drop across the 3 Ohm resistance is 8.01 Volts.
To calculate the potential drop across the internal resistance, we can use the same formula:
Voltage drop (Vdrop_internal) = I * Rinternal = 2.67 Amps * 1.5 Ohm = 4.005 Volts
So, the potential drop across the internal resistance is 4.005 Volts.
To calculate the internal voltage of the battery, we add the potential drop across the internal resistance to the external voltage drop:
Internal voltage (Vinternal) = E - Vdrop_internal = 12V - 4.005V = 7.995 Volts
Therefore, the internal voltage of the battery is 7.995 Volts.
In an open circuit, no current flows, so the voltmeter reads the entire emf of the battery. Therefore, the reading of the voltmeter connected across the battery terminals in an open circuit is 12 Volts.