22) Two 1.50 V cells, each with internal resistance of 0.0850 , are connected in

Question

22) Two 1.50 V cells, each with internal resistance of 0.0850 , are connected in series to form a battery that provides a current of 0.430 A to the external circuit. a. Find the current through each cell. b. What is the emf of the battery? c. What is the voltage applied to the external circuit? 23) In the circuit shown in the figure R two 360 emf A voltmeter with internal resistance of 635 Find the emf of the source. resistors are connected in series with an ideal source of is connected across one of the resistors and reads 3.23 V. FIGURE R

Explanation / Answer

a) As the batteries are in series the current must be same through each battery.
i.e 0.430 A

b) emf of the battery = 2*1.5

= 3.00 V

c) Terminal voltage of each battery = 1.5 - 0.085*0.43

= 1.463 A

so, the voltage applied to the external circuit = 2*1.463

= 2.93 V

23)

let V_emf is the emf of the source.

voltage across the voltmeter, V_volt = V_emf*(360*6350/(360 + 6350))/( 360*6350/(360 + 6350) + 360)

3.23 = V_emf*0.4862

V_emf = 3.23*0.4862

= 6.64 V

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