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

In the circuit shown in the figure, the current through:

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

3Ω resistor is 1.00 A

3Ω resistor is 0.25 A

4Ω resistor is 0.50 A

4Ω resistor is 0.25 A

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Concept: By using the formulas for resistance in series and resistance in parallel, we shall determine the equivalent resistances of each circuit one at a time. Then, using the present formula, i.e.

I = \frac{V}{R}

.
Formula used: I =

\frac{V}{R}

Total resistance =

\sum_{i = 1}^{n} R_{i}

And total resistance =

\frac{1}{\sum_{i = 1}^{n} \frac{1}{R_{i}}}

Let's first determine the equivalent resistance before calculating the current.
For series resistance-
Observe carefully that the resistance given in the last branch that is, 2Ω−4Ω−2Ω are in series. So, let the total resistance be

R_{1}

which will be-
⇒

R_{1}

=2+4+2
⇒

R_{1}

=8Ω
Now, this

R_{1}

is parallel to the other 8Ω and 8Ω. So, the equivalent resistance

R_{2}

will be-
⇒

R_{2}

\frac{1}{\frac{1}{8} + \frac{1}{8}}

⇒

R_{2} = 4 Ω

Now, as we can see that this is in series

R_{2}

with the 2Ω and 2Ω. The comparable resistance,

R_{3}

, will thus be-
⇒

R_{3}

= 2+4+2
⇒

R_{3}

= 8Ω
Now, again

R_{3}

is parallel to the next 8Ω resistance.
So, the resistance

R_{4}

will be
⇒

R_{4}

\frac{1}{\frac{1}{8} + \frac{1}{8}}

⇒

R_{4} = 4

Ω
Now, this R4 is in series with the next 3Ω and 2Ω. They are also forming the final equivalent resistance. Consequently, the ultimate comparable resistance will be
⇒equivalent resistance= 3+4+2
⇒equivalent resistance = 9 Ω
As we know that current is calculated by