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A battery whose emf is 40 $\mathrm{Vhas}$ internal resistance of 5$\Omega .$ If this battery is connected to a 15$\Omega$ resistor $R,$ what will the voltage drop across $R$ be?(A) 10 $\mathrm{V}$(B) 30 $\mathrm{V}$(C) 40 $\mathrm{V}$(D) 50 $\mathrm{V}$

Physics 102 Electricity and Magnetism

Chapter 11

Direct Current Circuits

Direct-Current Circuits

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Lectures

11:53

In physics, a charge is a …

10:30

The electric force is a ph…

00:36

A battery whose emf is 40 …

03:04

Thermal energy is to be ge…

01:56

(a) What is the current in…

03:25

(a) Find the current in an…

03:59

A wire of resistance 5.0$\…

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Consider the circuit shown…

01:04

What voltage is applied to…

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A $39-\Omega$ resistor is …

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A $24-\mathrm{V}$ emf is c…

0:00

In Fig. $27-50,$ two batte…

Hi in the given problem, E M F electro motive force of the battery Is E. is equal to 40 volt. And internal resistance of the battery is R is equal to five mm external resistance which is attached with the same battery is given as R is equal to 15 home. And as we know, external resistance is always in series with the internal resistance. So the electric current passing through the circuit will be given by the ratio of E. M. F. Of the battery to the net resistance offer means a plus up. So here this I will come out to be 40 divided by 15 plus five means this is 40 by 20 means this is 2.0 ampere hands. The voltage drop taking place across the external resistance are will be given by own slope as we is equal to I into our So this is 2.0 ampere multiplied by 15 ohm. And finally this potential drop Taking place across the external resistance are comes out to be 30 volt. So we can see here. A good option. B is correct. Thank you.

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