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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
⇒ I = 6 / 6 = 1 A
Hence, current in 6 Ω resistor is 1 A