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A 6-volt battery of negligible internal resistance is connected across a uniform wire AB of length 100cm. The positive terminal of another battery of emf 4V and internal resistance 1(Omega)is joined to the point A as shown in figure.Take the potential at B to be zero. (a) What are the potentials at the poits A and C ? (b) At which point D of the wire AB, the potential is equal to the potential at C?(c ) If the point C and D are connected by a wire, what will be the current through it ? (d) If the 4V battery is replaced by 7.5V battery,wht would be the answer of parts (a) and (b) ? |
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Answer» SOLUTION :Potential difference between AB is 6V. B is at 0 potential. Thus potential of a point is 6V. The potential difference between AC is 4V. `V_A- V_C = 4 ` ` V_C = V_A -4 = 6-4 = 2V. ` (b) The potential difference at D =` 2V = V_AD = 4V`. ` V_BD = 0V ` Current through the resisters R_1 & R_2 are equal. Thus , ` 4/R_1 = 2/R_2 ` ` rArr R_1/R_2 = 2 ` ` rArr l_1/l_2 = 2 ` (Acc.to the law of potentiometer) ` rArr l_1 + l_2 = 100 cm ` ` rArr l_1 + (l_1/2) = 100 cm ` ` rArr 3l_1/2 = 100cm ` ` rArr = l_1 = 200/3 cm = 66.67 cm .` ` AD = 66.67 cm` (c)When the points C and D are connected by a wire, current FLOWING through it is 0, since the points are equipotential . (d) Potential at A = 6V. ` Potential at C = 6 - 7.5 =- 1.5 V. ` The potential at B= 0 and towards A potential INCREASES. Thus negative potential point does not come within the wire. |
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