KVPY SB SX Exam 2011 Question Paper

© examsiri.com

Question : 99 of 120

Marks: +1, -0

Two batteries

V_1

and

V_ 2

are connected to three resistors as shown below.

If

V_1= 2 V

and

V_2= 0 V

. the current I = 3 mA. If

V_1= 0 V

and V

_2 = 4 V

. the current I = 4 mA. Now. if

V_1 = 10 V

and

V_ 2 = 10 V

, the current

I

will be

7 mA
15 mA
20 mA
25 mA

Solution:

{V}_{ {eq}}=\;{\;{ {V}_{1}}/{ {R}_{1}}+{ {V}_{2}}/{ {R}_{2}}}/{{1}/{ {R}_{1}}+\;{1}/{ {R}_{2}}} ⇒ \;{ {V}_{1} {R}_{2}+ {V}_{2} {R}_{1}}/{ {R}_{1}+ {R}_{2}}

{R}_{ {eq}}=\;{ {R}_{1} {R}_{2}}/{ {R}_{1}+ {R}_{2}}

{I}=\;{ {V}_{ {eq}}}/{ {R}+ {R}_{ {eq}}}

In each case

R_{e q} \& R

is same only

{V}_{1} \& {V}_{2}

is changing

∴ {V}_{ {eq}}

is changing

{V}_{ {eq}}=\;{2 × {R}_{2}+0 × {R}_{1}}/{ {R}_{1}+ {R}_{2}} & {[ {V}_{1}=2, {V}_{2}=0]}

{V}_{ {eq}}=\;{2 {R}_{2}}/{ {R}_{1}+ {R}_{2}}

Case - 2

\;\; {V}_{ {eq}}=\;{4 {R}_{1}}/{ {R}_{1}+ {R}_{2}} \;\; [ {V}_{1}=0, {V}_{2}=4]

\;{ {I}_{1}}/{ {I}_{2}}=\;{3}/{4}=\;{2 {R}_{2}}/{4 {R}_{1}} \;\; \;{ {R}_{2}}/{ {R}_{1}}=\;{3}/{2}

Case - 3

\;\; {V}_{ {eq}}=\;{10 {R}_{1}+10 {R}_{2}}/{ {R}_{1}+ {R}_{2}}

\;{3}/{ {I}^{'}}=\;{2 {R}_{2}}/{10( {R}_{1}+ {R}_{2})} ⇒ \;{3}/{ {I}^{'}}=\;{2 × 1.5 {R}_{1}}/{10(2.5 {R}_{1})}

or

{I}^{'}=25 {mA}

© examsiri.com

Go to Question:

Prev Question

Next Question