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A Battery of E.M.F. 15 V and Internal Resistance 2 Omega is Connected to Tvvo Resistors of 4 Ohm and 6 Ohm Joined. - Physics

Numerical

A battery of e.m.f. 15 V and internal resistance 2 `Omega` is connected to tvvo resistors of 4 ohm and 6 ohm joined.
(i) In series,
(ii) In para 1 lel. Find in each case the electrica I energy spent per minute in 6 ohm resistor.

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Solution

Given, emf= 15 V, internal resistance, r= 2 `Omega`

(i) Total resistance of given resistors in series, Rs = 4 + 6 = 10 `Omega`

Now, current through series combination on, I= `"emf"/("R"_"s" + "r") = 15/(10 + 2) = 15/12` = 1.25 A

In series, the same current will pass through each resistor

.·. electrical energy spent per minute in the 6`Omega` resistor, H= I(R) t = (1.25)(6) (60) = 562.5 J

(ii) Total resistance of given resistors in parallel, R= `(1/4 + 1/6) ^-1 = 2.4 Omega`

Current in parallel circuit, I = `15/(2.4 + 20) = 3.4 "A"`

In parallel, each resistor is connected across the same voltage, say V.

Then, V = e - Ir= 15 - (3.4 x 2) = 8.18 volt

:. electrical energy spent per minute in the 6Ω resistor, H= `"V"^2/"R" t = (8.18)^2/6 xx 60 `= 669.1 J

  Is there an error in this question or solution?
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APPEARS IN

Frank ICSE Class 10 Physics Part 2
Chapter 4 Current Electricity
Exercise 4.2 | Q 23 | Page 188
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