 Physics(a) Given, e.m.f. = 15 V internal resistance r = 3 Ω current through battery = ? If Rp is the equivalent resistance of resistors 3 Ω and 6 Ω connected in parallel, then 1Rp=13+161Rp=2+161Rp=361Rp=12⇒Rp=2Ω\dfrac{1}{Rp} = \dfrac{1}{3} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{2 + 1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{3}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1}{2} \\[0.5em] \Rightarrow R_p = 2 ΩRp1=31+61Rp1=62+1Rp1=63Rp1=21⇒Rp=2Ω From relation, ε = I (R + r) Substituting the value in the formula above we get, 15 = I(2 + 3)⇒ 15 = I x 5 ⇒ I = 15 / 5 = 3 A Hence, current through the battery = 3 A (b) Potential difference between the terminals of the battery = ? Using Ohm's law V = IR R = 2 Ω I = 3 A Substituting the values in the formula above we get, V = 3 x 2 = 6 V Hence, potential difference between the terminals of the battery = 6 V (c) Current in 3 Ω resistor = ? Using Ohm's law V = IR R = 3 Ω V = 6 V I = ? Substituting the values in the formula above we get, 6=I×3I=63⇒I=2A6 = I × 3 \\[0.5em] I = \dfrac{6}{3} \\[0.5em] \Rightarrow I = 2 A6=I×3I=36⇒I=2A Hence, current in 3 Ω resistor is 2 A (d) Current in 6 Ω resistor = ? Using Ohm's law V = IR R = 6 Ω V = 6 V I = ? Substituting the values in the formula above we get, 6 = I × 6 Hence, current in 6 Ω resistor is 1 A |
Two resistors of resistance 3 ohm and 6 ohm are connected in parallel to a cell of emf 1.5 V
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