The emf of a battery is 9 volt when 2 resistors
of 2 ohm and 3 ohm are connected in series
with the battery current drawn is 1.5 A
What will be the magnitude of the current
drawn from it if the resistors are joined
with battery in parallel?
Answers
Explanation:
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=
R
V
=
12A
15V
=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