In the arrangement shown in figure, the current through $$5\Omega$$ resistor is?
In this circuit there are $$2$$ cells connected in parallel
Equivalent internal resistance , $$r_{eq}= \dfrac{r_1 +r_2}{r_1* r_2}$$
where $$r_1$$ and $$r_2 $$ are the respective internal resistance
hence
$$r_{eq}= 1 \Omega$$
And Equivalent e.m.f, $$\dfrac{E_{eq}}{r_{eq}}= \dfrac{E_{1}}{r_{1}}+ \dfrac{E_{2}}{r_{2}}$$
Hence on solving we get
$$E_{eq}= 12 V$$
Hence current through $$5 \Omega$$ resistor
$$I= \dfrac{ E_{eq}}{r_{eq}+5}= \dfrac{12}{1+5}= 2 A$$
A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f 200 V through a resistance of $$1\Omega.$$ The battery terminals are connected to an external resistance 'R'. The minimum value of 'R', so that a current passes through the battery to charge it is :
In the given circuit diagram, the currents, $$I_1 = -0.3 A, I_4 = 0.8 A$$ and $$I_5 = 0.4 A$$ are flowing as shown. The currents $$I_2I_3$$ and $$I_6$$ respectively, are:
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Calculate I for the given circuit diagram.
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Find the magnitude and direction of the current flowing through the resistance $$R$$ in the circuit shown in Fig. if the emf's of the sources are equal to $$\varepsilon_{1} = 1.5\ V$$ and $$\varepsilon_{2} = 3.7\ V$$ and the resistances are equal to $$R_{1} = 10\Omega, R_{2} = 20\Omega, R = 5.0\Omega$$. The internal resistances of the sources are negligible.
Find the current flowing through the resistance $$R$$ in the circuit shown in Fig. The internal resistances of the batteries are negligible.
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