Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

What would be the reading of the ammeter when key ‘K’ is closed. If three identical resistors are connected in parallel with a cell and the reading of the ammeter is $0.6 A$ when key ‘K’ is open as shown in the figure below?

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

0.9 A

0.8 A

0.7 A

0.6 A

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

0.9 A

is the ammeter when key K is closed. If three identical resistors are connected in parallel with a cell and the reading of the ammeter is

0.6 A

when key K is open.
Let

R_{1} = resistance of 1 resistor

R_{2} = resistance of 2 resistor

When key 'K' is pressed, there are only two resistances in the circuit, hence the resultant resistance (

R_{P}

) is the parallel combination of two resistances.

\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

\frac{1}{R_{P}} = \frac{1}{R} + \frac{1}{R}

\frac{1}{R_{P}} = \frac{2}{R}

= > R_{P} = \frac{R}{2}

Let potential difference of cell

= V

volt
Current

I = 0.6 A

(given)

∴ V = I R_{P}

V = 0.6 \times \frac{R}{2}

V = 0.3 R

…………………………. (i)
Now when key 'k' is pressed, all three resistances in the circuit become parallel. Therefore the net resistance (

{R'}_{P}

) is calculated by using formula,

\frac{1}{{R'}_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

\frac{1}{{R'}_{P}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R}

\frac{1}{{R'}_{P}} = \frac{3}{R}

= > {R'}_{P} = \frac{R}{3}

…………………….. (ii)
According to ohm’s formula,

V = I {R'}_{P}

Now put the values of equations (i) and (ii) in above equation

= > 0.3 R = I (\frac{R}{3})

= > I = 0.3 R \times \frac{3}{R}

I = 0.9 A

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!