A cell of emf 10 text{ V} and internal resist | vedclass

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(B) The circuit consists of three cells connected in parallel. Let the emfs be $\varepsilon_1 = 10 	ext{ V}$,$\varepsilon_2 = 7 	ext{ V}$,and $\vare

Vedclass · Accepted Answer

$E=2 	ext{ V}, r=0.4 \Omega$

Vedclass · Answer

(B) The circuit consists of three cells connected in parallel. Let the emfs be $\varepsilon_1 = 10 	ext{ V}$,$\varepsilon_2 = 7 	ext{ V}$,and $\varepsilon_3 = 20 	ext{ V}$,with internal resistances $r_1 = 3 \Omega$,$r_2 = 0.6 \Omega$,and $r_3 = 2 \Omega$.Since the third cell is connected with reversed polarity relative to the first two,the equivalent emf $E_{	ext{eq}}$ and equivalent internal resistance $r_{	ext{eq}}$ are given by:$\frac{E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} - \frac{\varepsilon_3}{r_3}$$\frac{E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{10}{3} + \frac{7}{0.6} - \frac{20}{2} = \frac{10}{3} + \frac{35}{3} - 10 = \frac{45}{3} - 10 = 15 - 10 = 5 	ext{ A}$For parallel combination,the equivalent internal resistance is:$\frac{1}{r_{	ext{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{3} + \frac{1}{0.6} + \frac{1}{2} = \frac{1}{3} + \frac{5}{3} + \frac{1}{2} = 2 + 0.5 = 2.5 	ext{ S}$$r_{	ext{eq}} = \frac{1}{2.5} = 0.4 \Omega$Now,$E_{	ext{eq}} = \left(\frac{E_{	ext{eq}}}{r_{	ext{eq}}}ight) 	imes r_{	ext{eq}} = 5 	imes 0.4 = 2 	ext{ V}$Thus,$E = 2 	ext{ V}$ and $r = 0.4 \Omega$.

Similar Questions

There are a large number of cells available,each marked $(6 \,V, 0.5 \,\Omega)$,to be used to supply current to a device of resistance $0.75 \,\Omega$,requiring $24 \,A$ current. How should the cells be arranged so that power is transmitted to the load using the minimum number of cells?

For two cells having the same $EMF$ $E$ and internal resistance $r$,the current passing through an external resistor of $6 \ \Omega$ is the same when both cells are connected either in parallel or in series. The value of internal resistance $r$ is . . . . . . $ \ \Omega$.

Assertion : $A$ larger dry cell has higher $emf$.
Reason : The $emf$ of a dry cell is proportional to its size.

$A$ total of $100$ cells,each having an $emf$ of $5\,V$ and an internal resistance of $1\,\Omega$,are connected to an external resistance of $25\,\Omega$ to obtain the maximum current. If each row contains an equal number of cells,what should be the number of rows?

The circuit in the figure shows two cells connected in opposition to each other. Cell $E_1$ has an $emf$ of $6 \ V$ and an internal resistance of $2 \ \Omega$. Cell $E_2$ has an $emf$ of $4 \ V$ and an internal resistance of $8 \ \Omega$. Find the potential difference between the points $A$ and $B$.

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