In the circuit below, the batteries have negligible internal resistance. The ide

Question

In the circuit below, the batteries have negligible internal resistance. The ideal ammeter A reads current exactly and the ideal voltmeter V reads voltage exactly. When the Switch.S' is open, the voltmeter reads 15.0 V. What is the voltage of the battery labelled with a question mark? When the Switch S is closed, what is the current through the ammeter A? When the Switch S is closed, what is the voltage read by voltmeter V? When the Switch S is closed, what is the power provided by each of the two batteries? When the Switch S is closed, what is the power dissipated by each of the resistors?

Explanation / Answer

Voltmeter reads 15 volt

That is , the current through 50 ohm resistor = 15/50 = 0.30 A

The current through 30 ohm resister = 0.30 ohm

The voltage difference across 75 ohm resister = 15 + 30*0.3 = 15+ 9 = 24 volt

Henece current through 75 ohm resister = 24 / 75 = 0.32 A

So current through   battery = 0.32 + 0.3   = 0.62 A

(a.) The value of emf E = 24 + 20*0.62 = 36.4   volt

(b.) Voltage difference across 50 ohm resister = 25 volt.

Current = 25 / 50   = 0.5 A

(c.)   Voltmerer reading = 25 volt.

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