(a)find the potential drops across the two resistors shown in figure (a). (b) A voltmeter of resistance `(600 Omega)`is used to measure the potential drop across the `(300 Omega)` resistor.What will be the measured potential drop?

Explanation

(a)the current in the circuit is `(100V)/(300 Omega+200 Omega)=0.2A.`
The potential dorp across the `(300 Omega)` resistor is
`(300 Omega)xx0.2A=60 V.`
Similarly, the drop scross the `(200 Omega)` resistor is 40 V.
(b) The equivalent resistance, when the voltmeter is connected across `(300 Omega)`,is
`R=(200 Omega)+(600 Omegaxx300 Omega)/(600 Omega+300 Omega)=(400 Omega).`
Thus, the main current from the battery is
`i=(100 V)/(400 Omega)=0.25 A.`
The potential drop across the `(200 Omega)` resistor is, therefore, `(200 Omegaxx0.25A=50 V)`and that across `(300 Omega)`is also 50 V. this is also the potential drop across the voltmeter is 50V.