In the given circuit,calculate $(i)$ total resistance of the circuit,and (ii) current shown by the ammeter.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $(i)$ Since $R_{1}$ and $R_{2}$ are in series,their resultant resistance,
$R_{S} = R_{1} + R_{2} = 3 + 2 = 5 \Omega$
Further,$R_{S}$ and $R_{3}$ are in parallel,their resultant resistance is given by
$R_{P} = \frac{R_{S} \times R_{3}}{R_{S} + R_{3}} = \frac{5 \times 5}{5 + 5} = 2.5 \Omega$
(ii) Also,the current $I$ shown by the ammeter is given by Ohm's law:
$I = \frac{V}{R_{P}} = \frac{2.5 \text{ V}}{2.5 \Omega} = 1 \text{ A}$

Explore More

Similar Questions

$(a)$ What is an electric circuit?
$(b)$ Calculate the number of electrons that flow per second to constitute a current of one ampere. The charge on an electron is $1.6 \times 10^{-19} \ C$.
$(c)$ Draw an electric circuit for studying Ohm's law. Label the circuit components used to measure electric current and potential difference.

How is an ammeter connected in a circuit?

What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads $5\, A$. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Potential and potential difference are vector quantities.

You are given three equal resistors. The number of resistances which can be obtained by joining them in series and in parallel grouping is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo