V ses tau=(V sos)/(VuIs+I)+(Vus+I)/(V soo)

The resistance R=(V)/(I) , where V=500 pm 5V and I=10pm0.5A . Find the percentage error in R.

The resistance R=(V)/(I) , where V=100 pm 5V and I=10pm0.2A . Find the percentage error in R.

The resistance R= (V)/(I) , where V= (100+-5.0) V and I=(10+-0.2)A . Find the percentage error in R .

The resistance R=(V)/(I) where V=(200+-5)V and I=(20+-0.2)A ,the percentage error in the measurement of "R" is :

The resistance R= (V)/(I) , where V= (100+-5.0) V and I=(10+-0.2)A . Find the percentage error in R .

The resistance R= (V)/(I) , where V= (100+-5.0) V and I=(10+-0.2)A . Find the percentage error in R .

The resistance R= (V)/(I) , where V= (100+-5.0) V and I=(10+-0.2)A . Find the percentage error in R .

The resistance R= (V)/(I) , where V= (100+-5.0) V and I=(10+-0.2)A . Find the percentage error in R .

In Fig.23, O A\ a n d\ O B are opposite rays: (i) If ( i i ) (iii) x=( i v ) (v) 25^(( v i )0( v i i ))( v i i i ) , (ix) (x) what is the value of ( x i ) (xii) y ? (xiii) (xiv) (xv) If ( x v i ) (xvii) y=( x v i i i ) (xix) 35^(( x x )0( x x i ))( x x i i ) , (xxiii) (xxiv) what is the value of ( x x v ) (xxvi) x ? (xxvii) (xxviii)

In the circuit shown in the figure, the input voltage V_(i) is 20V, V_(BE)=0 and V_(CE) = 0 . The values of I_(B) , I_(C) and beta are given by