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(D) The total $emf$ of the circuit is $E + E = 2E$. The total resistance of the circuit is $R + R_1 + R_2$. Thus,the current in the circuit is $i = \f

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$R = R_2 - R_1$

Vedclass · Answer

(D) The total $emf$ of the circuit is $E + E = 2E$. The total resistance of the circuit is $R + R_1 + R_2$. Thus,the current in the circuit is $i = \frac{2E}{R + R_1 + R_2}$. For the source with internal resistance $R_2$,the potential difference $V$ is given by $V = E - i R_2$. Given that $V = 0$,we have $E - i R_2 = 0$,which implies $E = i R_2$. Substituting the value of $i$,we get $E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$. Dividing both sides by $E$,we get $1 = \frac{2 R_2}{R + R_1 + R_2}$. Therefore,$R + R_1 + R_2 = 2 R_2$,which simplifies to $R = R_2 - R_1$.

Point	Slope of the normal
$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

Column $I$	Column $II$
$(a)$ Tidal volume	$(i)$ $2500 - 3000 \ mL$
$(b)$ Inspiratory Reserve volume	$(ii)$ $1100 - 1200 \ mL$
$(c)$ Expiratory Reserve volume	$(iii)$ $500 - 550 \ mL$
$(d)$ Residual volume	$(iv)$ $1000 - 1100 \ mL$

Two sources of equal $emf$ are connected to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

Similar Questions

Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.
Point Slope of the normal
$A. (7, 2)$ $1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$ $2. -8$
$C. (1, -1)$ $3. 4$
$D. (3, \sqrt{2})$ $4. 0$
$5. -2\sqrt{2}$

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