Name: Ohm’s Law & Introduction to Electric Resistance
Uploaded: 2019-07-19
Duration: 8 min 43 s
Description: In our attempt to establish a relationship between current and voltage, we will discover a new physical quantity in this video. Let's watch and learn.

Question

Find the voltmeter reading and ammeter reading in each of the following cases.

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Accepted Answer

(a) From the figure (a) , it is clear that the $5 Ω$ and $4 Ω$ resistors are in series connection. Therefore, the can be replaced by a single resistor with effective resistance $R_{S}$ .

Therefore, $R_{S} = 4 + 5 = 9 Ω$

Similarly, $2 Ω$ resistors are in parallel connection and can be replaced by a single resistor of resistance $R_{P}$ .

Therefore, $R_{P} = \frac{2 \times 2}{2 + 2} = 1 Ω$

Now the total equivalent resistance, $R_{T} = R_{S} + R_{P} = 10 Ω$

Given that voltage, $V = 10 V$

Therefore the current in the circuit, $I = \frac{V}{R_{T}} = \frac{10}{10} = 1 A$

So the ammeter will show a reading of $1 A$ .

The voltmeter is connected across the series combination of $4 Ω and 5 Ω$ resistors.

Therefore the voltmeter reading $V' = I \times R_{S} = 1 \times 9 = 9 V$ .

(b) From the figure (b) the $3 Ω$ resistors are in series connection and the combination is in parallel connection with $6 Ω$ resistor.

Therefore the equivalent resistance, $R_{e} = \frac{(3 + 3) \times 6}{(3 + 3) + 6} = 3 Ω$

Given that voltage, $V = 10 V$

Therefore the current in the circuit, $I = \frac{V}{R_{e}} = \frac{10}{3} = 3.33 A$

Therefore the ammeter will show a reading of $3.33 A$ .

The voltmeter is connected to one of the $3 Ω$ resistors. Since both the parallel circuits have equal resistance, the current will be divided equally and the $\frac{3.33}{2} A$ current passes through each path.

Therefore the voltmeter reading, $V'' = \frac{3.33}{2} \times 3 Ω = 5 Ω$

Find the voltmeter reading and ammeter reading in each of the following cases.

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Important Questions on Electricity

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