A real battery of emf and internal resistance is connected to the resistor network…

PHYSICS · Multi-part · CBSE · Physics · Current Electricity · Current Electricity

Problem

A real battery of emf

\mathcal{E}=9.0\,\text{V}

and internal resistance

r=1.0\,\Omega

is connected to the resistor network described below. • Resistors

R_{5}=12\,\Omega

and

R_{6}=24\,\Omega

are in series. • This series pair is connected in parallel with a single resistor

R_{7}=18\,\Omega

between terminals

X

and

Y

. • The battery is connected so that current leaves the positive terminal and enters terminal

X

, then exits terminal

Y

and returns to the negative terminal of the battery through the internal resistance. An ammeter of negligible resistance is placed in series between the battery and terminal

X

. A voltmeter of finite resistance

R_{v}=10\,\text{k}\Omega

is later connected across terminals

X

and

Y

. (1) Determine the equivalent resistance of the external resistor network between

X

and

Y

. Show all steps. (2) Calculate the current that the battery delivers to the external circuit when only the ammeter is connected (no voltmeter present). (3) Using the current found in part (b), compute the terminal potential difference

V_{XY}

across the battery when no voltmeter is present. (4) The voltmeter of resistance

R_{v}

is now connected across terminals

X

and

Y

. Without detailed calculation, explain qualitatively whether the reading of the voltmeter will be greater than, less than, or equal to the value of

V_{XY}

found in part (c). Support your answer with appropriate reasoning about how the voltmeter alters the circuit.

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