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Question: When terminals of a cell of emf \[1.5V\] are connected to an ammeter of resistance \(4\Omega \), the...

When terminals of a cell of emf 1.5V1.5V are connected to an ammeter of resistance 4Ω4\Omega , the ammeter reads 0.30A0.30A. Which of the following statements are correct?
A) The internal resistance of the cell is zero.
B) If a 4Ω4\Omega resistance is also across the terminals of the cell, the ammeter will read 0.50A0.50A .
C) If a 4Ω4\Omega resistance is also across the terminals of the cell, the ammeter will read 0.40A0.40A .
D) Suppose a voltmeter of resistance 4Ω4\Omega is used to measure the potential difference between terminals of a cell, it will read 1.0V1.0V.

Explanation

Solution

The ammeter is used to measure the current in the circuit and always connected in series with the external resistance, whereas voltmeter is used to measure the potential difference across an external resistance and always connected in parallel.

Complete step by step answer:
A cell is a single electrical energy source which uses chemical reaction to produce current.
When a current produce in a cell due to chemical reactions then a factor that’s opposing to the flow of current in cell is called internal resistance
The EMF (Electromotive force) of cell
EMF of cell =1.5V=1.5V
Resistance of ammeter =4Ω=4\Omega
Resistance of circuit =Vi=1.50.3=5Ω=\dfrac{V}{i}=\dfrac{1.5}{0.3}=5\Omega
Now the ammeter provides 4Ω4\Omega resistance only, so the surplus is coming from the cell.
The internal resistance of the cell is =54=1Ω=5-4=1\Omega
If a 4Ω4\Omega resistance is connected across cell (assuming parallel connection) resistance of the circuit will be=4×44+4+1=3=\dfrac{4\times 4}{4+4}+1=3
The final Emf across the cell will be
Eir=1.5(1.53)×1=1VE-ir=1.5-\left( \dfrac{1.5}{3} \right)\times 1=1V
Current through ammeter =14=0.25A=\dfrac{1}{4}=0.25A
When a voltmeter is connected to across the cell, the final resistance will be
4×44+4\dfrac{4\times 4}{4+4} (resistance of circuit)+1+1 (internal resistance of circuit)
=168+1 =2+1 =3Ω \begin{aligned} & =\dfrac{16}{8}+1 \\\ & =2+1 \\\ & =3\Omega \\\ \end{aligned}
The reading of the voltmeter
=Eir =1.5(1.53)×1 =1volt \begin{aligned} & =E-ir \\\ & =1.5-\left( \dfrac{1.5}{3} \right)\times 1 \\\ & =1volt \\\ \end{aligned}
Voltmeter resistance 4Ω4\Omega is used to measure the potential difference between terminals of the cell, it will be read 1volt1volt.

\therefore D is the correct option.

Note: The voltage provided/emf of an ideal battery and the one with internal resistance are different. Always connect the external resistance across the battery in parallel as otherwise the circuit would be incomplete.