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AS and A Level: Electrical & Thermal Physics

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Doing circuit calculations

  1. 1 To find the total resistance of a circuit follow these steps.

    1) Replace any parallel network with a single equivalent resistor, REQ using 1/REQ= 1/R1 + 1/R2.

    Tip: REQ will be lower than either of the parallel resistors R1 or R2 so you can check your calculation.

    2) Add all of the series resistors together (including REQ) to find the total resistance of the circuit RT.
  2. 2 Calculate the total circuit current, IT using IT = V/RT. This current flows through all of the series resistors so the p.d. across each series resistor is given by V = IT R. The p.d. across any parallel network will be IT REQ.
  3. 3 A potential divider circuit consists of two resistors in series. Follow the same steps as above to find the p.d. across each resistor. Alternatively, R1/R2 =V1/V2 or V1 = V *R1/(R1 +R2) [V = supply voltage]
  4. 4 Which bulb is brightest?

    1) If two bulbs are in series, they have the same current. The brighter bulb is the one with greatest power, P. Use P = I2R. The bulb with largest R is brightest.

    2) If two bulbs are in parallel, they have the same p.d. across them. Use P=V2/R. The bulb with the lowest R has the highest power and is therefore brightest.

Resistivity

  1. 1 Use the correct units. If diameter is given in mm, convert to metres before calculating area, A. e.g. d = 1mm so r = 0.5mm = 0.5 x 10-3 m. So A = x (0.5 x 10-3)2 = 7.9 x 10-7 m2.
  2. 2 Typical questions involve proportions such as what happens to R if the diameter of the wire is doubled? Doubling the diameter would double the radius. Doubling the radius would quadruple the area. So the resistance would decrease to ¼ of the original resistance. The same argument explains why a thinner wire has a higher resistance.
  3. 3 Applications of resistivity:

    1) A rheostat is a resistor made by winding a wire around a cylindrical tube. A sliding contact changes the length of the wire carrying current and therefore changes the resistance, R.

    2) A strain gauge, has a resistance that increases when it is stretched because the wire from which it is made increases in length.

    3) The battery tester on the side of some AA batteries works by using a shaped conductor. The thin end has lowest A, therefore highest R. Current is the same at all points, the thin end gets hottest (P = I2R) and a thermochromic ink becomes transparent, revealing a display.

Internal resistance

  1. 1 Many students find internal resistance a difficult concept. However the circuit is similar to a potential divider. Think of the circuit as a cell of emf E, in series with an internal resistance, r and an external resistance R. When current, I flows through the circuit, E = Ir + IR. This is Kirchhoff’s 2nd law.
  2. 2 Using a voltmeter to measure the terminal p.d. V, we can rewrite the equation E = Ir + IR as E = Ir + V and then rearrange to give V = rI + E which is the equation of a straight line. A graph of V against I gives a straight line of gradient -r and intercept E. This is how to find the emf experimentally.
  3. 3 When the current through the cell is high, there is a large drop in the terminal p.d. The difference between the cell emf and the terminal p.d. is called the ‘lost volts’ and equals Ir.
  4. 4 Short circuiting the cell will lead to a large drop in external voltage and large amount of power dissipated in the cell as P = I2r.
  5. 5 A car battery (lead acid) is designed to supply large currents. When switching on the engine the current is large and there will be a large drop in terminal p.d. and this will cause lights to dim momentarily.

  • Marked by Teachers essays 5
  • Peer Reviewed essays 9
  1. Diode Application in Rectifier Circuits

    The AC on each side of the center-tap is ½ of the total secondary voltage. Only one diode will be biased on at a time the prose to be center tap full-wave rectifier is like below. Last is about full-wave bridge rectifier circuit, The Bridge is shown in the figure below. The circuit has four diodes connected to form a bridge. The ac input voltage is applied to the diagonally opposite ends of the bridge. The load resistance is connected between the other two ends of the bridge.

    • Word count: 2206
  2. Energy Efficiency Experiments

    = 9171.998J Useful energy output= Mass x Specific heat capacity x Temperature change Useful energy output = 9.0414J Electrical energy → Heat energy Efficiency = 9171.998 ÷ 11940 x 100 = 77% Useful energy output ÷ energy input x 100 Energy input= Electrical. C:\Users\Hello\Pictures\TPhoto_00007.jpg Output = heat. Useful= heat. Wasted= heat. Energy→ heat. Method –Experiment 3 In this experiment we timed how long the glider went past the light gate. We measured the mass of the glider, height of the ramp at highest end, length of the ramp, measured the length of the card and recorded the time.

    • Word count: 1871

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