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Question: Find the reading of the ammeter in the given circuit? ![](https://www.vedantu.com/question-sets/8f...

Find the reading of the ammeter in the given circuit?

A. 0.40.4
B. 11
C. 0.60.6
D. 1.21.2

Explanation

Solution

Hint : An ammeter is an electrical instrument used to measure current.
The current through each resistor connected in series is the same and voltage across the resistors connected in parallel is the same.
According to Ohm's law, voltage across a resistor is given by V=IRV = IR where II is the current passing through the resistor and RR is the resistance.

Complete Step By Step Answer:
We know that an ammeter is an electrical instrument used to measure current.
We know that the equivalent resistance in a series combination of n resistors is given by Req=R1+R2+R3+.....+Rn{R_{eq}} = {R_1} + {R_2} + {R_3} + ..... + {R_n}
And the equivalent resistance in a parallel combination of n resistors is given by 1Req=1R1+1R2+1R3+.....+1Rn\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ..... + \dfrac{1}{{{R_n}}}
As clear from the figure that 10Ω10\Omega and 2Ω2\Omega resistances are in series so their equivalent resistance R1=10+2=12Ω{R_1} = 10 + 2 = 12\Omega
Similarly, 25Ω25\Omega and 5Ω5\Omega resistances are in series so their equivalent resistance R2=25+5=30Ω{R_2} = 25 + 5 = 30\Omega
Now, R1{R_1} and R2{R_2} will be in parallel and the total equivalent resistance of the circuit is given by
1Req=1R1+1R2=112+130\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} = \dfrac{1}{{12}} + \dfrac{1}{{30}}
On simplifying we have
Req=12×3012+30=8.57Ω{R_{eq}} = \dfrac{{12 \times 30}}{{12 + 30}} = 8.57\Omega
According to Ohm's law, voltage across a resistor is given by V=IRV = IR where II is the current passing through the resistor and RR is the resistance.
We know that the voltage across the resistors connected in parallel is the same.
So, the voltage across R1{R_1} will be given by
V=1.4×8.57=11.99 V12 VV = 1.4 \times 8.57 = 11.99{\text{ V}} \approx 12{\text{ V}}
Now, as we know that the current through each resistor connected in series is the same. So, the current through 10Ω10\Omega and 2Ω2\Omega resistances will be given by
I=VR1=1212=1 AI = \dfrac{V}{{{R_1}}} = \dfrac{{12}}{{12}} = 1{\text{ A}}
Therefore, reading of the ammeter is 1 A1{\text{ A}} .
Hence, option B is correct.

Note :
The Ohm’s law for current electricity is applicable only when the temperature and other physical condition remains constant. Always remember that the current through each resistor connected in series is the same and voltage across the resistors connected in parallel is the same.