Calculate the value of current I in given circuit .

2 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 .5 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 1 .5 A
REF.Image
in the given figure inclined wire is shorting the circuit,
and due to this shorting No current will flow through
Resister B,D,E,F (as marked in figure)
so for analysis parpose we can remove this resistor and
modified figure would be :
So Total resistance = $5 + 5 = 10 Ω$
so current $(I) = \frac{V}{R} = \frac{15}{10} = 1.5 A$

Suggest Corrections

Similar questions

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

For an increasing A.P. $a_{1}, a_{2}, a_{3} . . . a_{n}$ ,
if $a_{1} + a_{3} + a_{5} = - 12$ and $a_{1} \cdot a_{3} \cdot a_{5} = 80$ , then which of the following is true.

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

are in A.P. with common difference

\neq 0,

then find the value of

\sum_{i = 1}^{5} a_{i},

when

a_{3} = 2.

Q. If for an

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

and

a_{1} a_{2} a_{3} = 8

then the value of

a_{2} + a_{4} + a_{6}

equals

Q. For an increasing A.P.

a_{1}, a_{2}, a_{3}, \dots, a_{n},

a_{1} + a_{3} + a_{5} = - 12

and

a_{1} a_{3} a_{5} = 80

, then which of the following is (are) TRUE?

Join BYJU'S Learning Program