Physics, asked by orlie, 6 months ago

Two batteries of rating 10 V and 15 V are arranged with two resistors of 2ohm and
3ohm, two ammeters A, and A2 and two switches S and S2 are shown in figure.​

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Answers

Answered by Sanumarzi21
1

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

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Answered by dthdealer9
1

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=

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

Explanation:

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