Combine three resistors 5Omega- 4-5Omega and 3Omega in such a way that the total resistance of this combination is maximum-13-5Omega12-5Omega14-5Omega16-5Omega

Question

Toppr Admin · Accepted Answer

Given resistance-R1-5-x3A9-R2-4-5-x3A9-R1-3-x3A9-In series combination effective resistance will be maximum-ie- Req-R1-R2-R3-xA0- -xA0- -xA0-Req-5-x3A9-4-5-x3A9-3-x3A9-12-5-x3A9-Option A

Combine three resistors $5 Ω, 4.5 Ω$ and $3 Ω$ in such a way that the total resistance of this combination is maximum:
$13.5 Ω$
$12.5 Ω$
$14.5 Ω$
$16.5 Ω$

Given resistance,
$R_{1} = 5 Ω R_{2} = 4.5 Ω R_{1} = 3 Ω$

In series combination effective resistance will be maximum.
ie, $R_{e q} = R_{1} + R_{2} + R_{3}$
$R_{e q} = 5 Ω + 4.5 Ω + 3 Ω = 12.5 Ω$

Option A

Combine three resistors 5Ω,4.5Ω and 3Ω in such a way that the total resistance of this combination is maximum:13.5Ω12.5Ω14.5Ω16.5Ω

Given resistance,R1=5ΩR2=4.5ΩR1=3ΩIn series combination effective resistance will be maximum.ie, Req=R1+R2+R3 Req=5Ω+4.5Ω+3Ω=12.5ΩOption A

Combine three resistors $5 Ω, 4.5 Ω$ and $3 Ω$ in such a way that the total resistance of this combination is maximum:
$13.5 Ω$
$12.5 Ω$
$14.5 Ω$
$16.5 Ω$

Given resistance,
$R_{1} = 5 Ω R_{2} = 4.5 Ω R_{1} = 3 Ω$

In series combination effective resistance will be maximum.
ie, $R_{e q} = R_{1} + R_{2} + R_{3}$
$R_{e q} = 5 Ω + 4.5 Ω + 3 Ω = 12.5 Ω$

Option A