A cell of negligible internal resistance is connected
A cell of negligible internal resistance is connected in series with two resistors of 100Ω and 200Ω. The current through the resistors is 300 m. A. 1. Calculate the terminal p. d. of the cell. 2. What is the e. m. f. of the cell? Always draw the arrangement V 1. The p. d. across the terminals = the p. d. across the external resistors V=IR V = 300 x 10 -3 x 300 V= 900 x 10 -3 V = 90. 0 V 2. The question tells us that th internal resistance of the cell is negligible. I= 300 x 10 -3 A 100Ω 200Ω When this is the case The e. m. f. = terminal p. d. V E = 90. 0 V
The terminal potential of an unconnected cell is 1. 51 V. The cell is connected in series with a 10 kΩ and a microammeter of negligible resistance. The p. d. across the resistance is measured as 1. 49 volts by a high resistance voltmeter. Calculate the internal resistance of the cell: The terminal potential of the unconnected cell is the e. m. f. of the cell r The current through the series circuit is the same everywhere The resistance of this meter is negligible 10 kΩ 1. 49 V μA
r 10 kΩ 1. 49 V E = 1. 51 V μA Applying the formula for internal resistance:
V A R 1 R 2 A cell is connected to a switch and two resistors R 1 and R 2 with values of 100Ω and 400Ω which are in parallel. When the switch S is open, the high resistance voltmeter V reads 1. 63 V. When switch S is closed V reads 1. 61 V. 1. Calculate the current registered by the ammeter A assuming that it has negligible resistance. 2. Calculate the internal resistance of the cell
A cell is connected to a switch and two resistors R 1 and R 2 with values of 100Ω and 400Ω which are in parallel. When the switch S is open, the high resistance voltmeter V reads 1. 63 V. When switch S is closed V reads 1. 61 V. 1. Calculate the current registered by the ammeter A assuming that it has negligible resistance. The value of the resistance in the circuit is given by The current though the “series” circuit I = 0. 02 A
2. Calculate the internal resistance of the cell ε=1. 63 V. I = 0. 02 A 1. 63 = 0. 02 x 80+ 0. 02 r r = (1. 63 -1. 60)/0. 02 V A R 1 R 2
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