The series combination of two batteries both of the same emf 10 V , but different internal resistance of $20 Ω$ and $5 Ω$ is connected to the parallel combination of two resistors $25 Ω$ and $R Ω$ .The voltage difference across the battery of internal resistance $20 Ω$ is zero . The value of $R (in Ω)$ is,

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Infinity Learn · Accepted Answer

For $$10$$ V, $$20$$Ω battery : ∴  $$V1$$ $$=$$ ε$$1$$−$$ir1$$ ⇒$$0=10$$−i×$$20$$ $$i=0.5A$$  For $$10$$ V, $$5$$Ω battery : ∴ $$V2$$ $$=$$ ε$$2$$ −i $$r2$$ $$=10$$−$$0.5$$×$$5=7.5V$$  The potential difference across all three branches is equal. V $$=$$ $$V1$$ $$+$$ $$V2$$  $$=$$ $$0$$ $$+$$ $$7.5$$ $$=$$ $$7.5$$ V  Now, i $$=$$ $$i1$$ $$+$$ $$i2$$ ⇒$$0.5$$ $$=$$ $$7.525$$ $$+7.5R$$ ⇒$$0.5=0.3+7.5R$$ ⇒$$0.2=7.5R$$ ⇒R $$=$$ $$7.50.2=37.5$$ Ω

Sources of emf and Kirchhoff's laws

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Sources of emf and Kirchhoff's laws

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The series combination of two batteries both of the same emf 10 V , but different internal resistance of 20Ω and 5Ω is connected to the parallel combination of two resistors 25Ω and R Ω .The voltage difference across the battery of internal resistance 20 Ω is zero . The value of Rin Ω is,

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