5.3 Emf and internal resistance Internal resistance * is the opposition to charge moving through a source of electricity * causes electrical energy to be dissipated inside the source as charge moves through it Producing heat ! Electromotive force of a source * is the potential (chemical) energy transferred to electrical energy when one coulomb of charge passes through it E =W Q 5.3 Emf and internal resistance A high resistance voltmeter measures terminal Pd. E V r Open circuit Pd of a cell equals the EMF in volts, because the current through the cell is negligible. Energy supplied per coulomb by cell I E Energy wasted per coulomb by internal resistance = Energy changed per coulomb by external circuit EMF = Pd across R + Pd across r E = V + v E = IR + Ir E = I ( R + r) + R r 5.3 Emf and internal resistance Q. Find the internal resistance of a cell if its emf is 3V and the pd across an external resistance is 2.5 volts when 0.5 A flows. E V r EMF = Pd across R + Pd across r E = V + v “terminal pd” I E E = IR + Ir 3 = 2.5 + 0.5 r 0.5 = 0.5 r r = 1.0 ohm R r “lost volts” 5.3 Emf and internal resistance Power Power supplied by cell = IE = I(IR +Ir) = I 2 R + I 2r Useful power delivered to R Maximum power delivered to R Power delivered to R * occurs when the load resistance R matches the internal resistance r I2 R = E2 R (R+r)2 (R+r)I = I = E E (R+r) r Load resistor R 5.3 Emf and internal resistance Measurement of emf and internal resistance r * adjust the variable resistor for different values of current and record the terminal pd across the cell r R limits the current to safe levels ! Terminal Pd /V 2 Emf = 1.5V 1 0.5 1.0 Circuit current / A 5.3 Emf and internal resistance Measurement of emf and internal resistance r Terminal Pd /V 2 EMF = Pd across R + Pd across r E = V + v E = IR + Ir IR = E - Ir V = E - rI - mX Compare with straight line equation: Y Emf = 1.5V = c m = gradient (0, 1.5) (1.2 , 1) 1 m= Y Int resistance r = m = -0.5 = 0.42 0.5 1.0 Circuit current / A 1.2