📖 generic · CBSE Class 12th English Medium · PHYSICS PART-1 · Page 115question

C HARGES IN CLOUDS · Part 2

Chapter 3: Chapter 3 · PHYSICS PART-1

average electric field is about V/m, which corresponds to a surface charge density of – . × – C/m . Over the entire earth’s surface, the total negative charge amount to about kC. An equal positive charge exists in the atmosphere.

This electric field is not noticeable in daily life. The reason why it is not noticed is that virtually everything, including our bodies, is conductor compared to air. E XAMPLE . Example .

A network of resistors is connected to a V battery with internal resistance of Ω , as shown in Fig. . : (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor.

(c) Obtain the voltage drops V AB , V BC and V CD . FIGURE . Solution (a) The network is a simple series and parallel combination of resistors. First the two Ω resistors in parallel are equivalent to a resistor = [( × )/( + )] Ω = Ω.

In the same way, the Ω and Ω resistors in parallel are equivalent to a resistor of [( × )/( + )] Ω = Ω . The equivalent resistance R of the network is obtained by combining these resistors ( Ω and Ω ) with Ω in series, that is, R = Ω + Ω + Ω = Ω . (b) The total current I in the circuit is A ( ) I Ω Consider the resistors between A and B. If I is the current in one of the Ω resistors and I the current in the other, I × = I × that is, I = I , which is otherwise obvious from the symmetry of the two arms.

But I + I = I = A. Thus, I = I = A that is, current in each Ω resistor is A. Current in Ω resistor between B and C would

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