Two resistors of resistances R1=100+-3ohm and …

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Two resistors of resistances R1=100+-3ohm and R2=200+-4ohm are connected (a) in series, (b) in parallel.Find the equivalent resistance of the (a) series combination, (b) parallel combination.Use for (a) the relation R1=R1+R2 and for (b) 1/R'=1/R1+1/R2 and ∆R'/R'2=∆R1/R1^2+∆R2/R2^2

Posted by Yakub Saiyed 6 years, 4 months ago

CBSE > Class 11 > Physics

1 answers

Sia ? 6 years, 4 months ago

Here, R₁ = (100 ± 3) {tex}\Omega{/tex}, R₂ = (200 ± 4) {tex}\Omega{/tex}

Resistance in Series combination
R = R₁ + R₂ = 100 + 200 = 300 {tex}\Omega{/tex}
{tex}\Delta{/tex}R = ± ({tex}\Delta R _ { 1 } + \Delta R _ { 2 } {/tex})
= ±(3 + 4) = ± 7 {tex}\Omega{/tex}
{tex}\therefore \quad R = ( 300 \pm 7 ) \Omega{/tex}
Resistance in Parallel Combination
{tex}\frac { 1 } { R ^ { \prime } } = \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 2 } } = \frac { R_2 + R_1 } { R_1 R_2 } = \frac{200+100}{100 \times 200}{/tex}{tex} = \frac { 300} { 20000 } = \frac { 3 } { 200 }{/tex}
{tex}R ^ { \prime } = \frac { 200 } { 3 } {/tex}= 66.66667 {tex}= 66.7 \Omega{/tex} [ upto one decimal place]
{tex}\frac { \Delta R ^ { \prime } } { R ^ { \prime 2 } } = \pm \left[\frac { \Delta R _ { 1 } } { R _ { 1 } ^ { 2 } } + \frac { \Delta R _ { 2 } } { R _ { 2 } ^ { 2 } }\right]{/tex}
{tex}\Delta R ^ { \prime } = \pm \left[\Delta R _ { 1 } \left( \frac { R ^ { \prime } } { R _ { 1 } } \right) ^ { 2 } + \Delta R _ { 2 } \left( \frac { R ^ { \prime } } { R _ { 2 } } \right) ^ { 2 }\right]{/tex}
{tex}= \pm \left[3 \left( \frac { 200 } { 3 \times 100 } \right) ^ { 2 } + 4 \left( \frac { 200 } { 3 \times 200 } \right) ^ { 2 } \right]{/tex}{tex}= \pm \left[3 \left( \frac { 4 } { 9 } \right) + 4 \left( \frac { 1 } { 3 } \right) \right]{/tex}{tex}= \pm 1.8 \ \Omega{/tex}
Hence, R' = ( 66.7 ± 1.8 ) {tex}\Omega{/tex}