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Transcript
Seven - Series and Parallel Circuits 1. state that Kirchoff's second law is = the sum of the e.m.f.'s around a loop is equal to the sum of the p.d.'s around the same loop - appreciate that this is due to the conservation of energy 2. define series circuit as a circuit in which the components are connected end-to-end and therefore are in the same loop, so there is only one path for current to flow - p.d. is shared between them whereas current is constant 3. define parallel circuit as a circuit in which there is more than one loop connected to the power supply, therefore more than one path for current to flow -current is always shared between loops whereas p.d. remains constant π π’ππππ¦ π.π.π. - πππ ππ π‘ππππ = π‘ππ‘ππ ππ’πππππ‘ 4. prove and use the equation for resistances in series circuit: RTOT = R1 + R2... by applying KII: VTOT = V1 + V2 by applying KI: IRTOT = IR1 + IR2 cancel out I's because KI states that I is constant in same loop: RTOT = R1 + R2 also resistors in series means an increase in length of resistors and Rβlength, so increase in resistance 5. prove and use the equation for resistances in parallel circuits: π 1 πππ 1 1 =π +π +β― 1 2 by applying KI: ITOT = I1 + I2 by applying KII: VTOT = V1 = V2 V's get cancelled out from above equation overall, cross-sectional-area of resistors increase, and π β 1 π.π .π , so resistance decreases 6. solve circuit problems involving series and parallel circuits with more than one sources of e.m.f. E1 > E2 I3 = I1 + I2 E1 = I3 x R1 E2 = (I3 x R1) + (I2 x R2) 7. explain that all sources of e.m.f. have an internal resistance: - in battery due to chemicals - in power supplies due to the components inside (e.g. wires) 8. select and use the equations: - E = I (R + r) where R is the load resistance and r is the internal resistance and Ir is the "lost volts" - E = V + Ir 9. define terminal potential difference as the p.d. across the load (external) resistance - terminal p.d. is the energy transferred when I coulomb of charge flows through the external resistance - V = E - Ir 10. describe the experiment used to show internal resistance: set up circuit set power pack to 10V set resistance to low record I and V change R record new I and V repeat plot V against I compare to y = mx + c, y= V and x= I m = -r and c = e.m.f. 11. explain maximum power dissipation in the external resistance occurs when both internal and external resistances are equal, and when both resistances are equal for maximum power, the efficiency is 50% 12. explain that maximum current is when the external resistance is zero (short circuited) 13. explain that maximum voltage is when the external resistance is 'infinite' and the efficiency approaches 100% 14. explain that when the external resisitance is either zero, power dissipation in it is zero When the external resistance is zero, the power dissipation in the internal resistance is maximum (i.e. we have 0% efficiency) 15. understand that power that can be dissipated externally is calculated by: ππππ‘ππππΈπππΈπ ππ΄πΏ πππ€ππ = π ππ ππ π‘πππππΈπππΈπ ππ΄πΏ × ( )2 (π ππ ππ π‘πππππΈπππΈπ ππ΄πΏ + π ππ ππ π‘πππππΌπππΈπ ππ΄πΏ )