The above figure shows a typical no-load saturation curve or open circuit characteristics for all types of DC generators. 2. Internal or Total Characteristic (E/I a) An internal characteristic curve shows the relation between the on-load generated emf (Eg) and the armature current (I a).
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Jul 01, 2014 · Two cells of emf E 1 and E 2 (E 1 > E 2) are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. Calculate the ratio of E 1 and E 2.


  • The figure shows three identical lightbulbs connected to a battery having a constant voltage across its terminals. What happens to the brightness of lightbulb 1 when the switch S is closed? A is the brightest, and B and C have equal brightness but less than A.
  • 11. Bulbs A, B, C, and D are identical and connected to a battery as shown. The bulb with the least current passing through it is A. bulb A. B. bulb C. ... 36. If the resistance in a circuit using an ideal (no internal resistance) battery is increased, then ... A resistor R is connected across a battery of negligible internal resistance. If the ...

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A series connection of two voltage sources. The emfs (each labeled with a script E) and internal resistances add, giving a total emf of emf 1 + emf 2 and a total internal resistance of r 1 + r 2. Figure 9. Batteries are multiple connections of individual cells, as shown in this modern rendition of an old print.

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Mar 02, 2011 · a circuit consisting of six identical resistors, each of resistance R = 10 kO, a capacitor of capacitance C = 50 µF and a battery supplying a potential difference of Vbatt = 2.0 V. You may assume that the internal resistance of the battery is negligible. Figure 3 Circuit diagram for Question 3.

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Jul 11, 2010 · The brightness of the bulbs is determined by the amount of the electric current flowing through them. So, in order to decide what will happen to the brightness of the bulbs B and C if bulb A burns ...

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A battery has an emf of 12.0 V and an internal resistance of 0.05 Ω. Its terminals are connected to a load resistance of 3.00 Ω. Find the terminal voltage of the battery.

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2. In the circuit shown below, what is the p.d across the bulb, and the switch when: I. The switch is open? II. The switch is closed? 3. Define a volt 4. Two cells, A and B connected with parallel are in series with a bulb as shown below. I. Copy the diagram to show where the ammeter should be connected in order to measure the current through ...

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The circuit shown consists of just two resistors, R 1 and R 2 connected together in series across the supply voltage V in. One side of the power supply voltage is connected to resistor, R 1, and the voltage output, V out is taken from across resistor R 2. The value of this output voltage is given by the corresponding formula.

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In the network shown in Fig. 13.50, the battery. has negligible internal resistance. Find, using Th´evenin’s and Norton’s theorem, the current flowing in the 4 Ω resistor. [0.918 A] Figure 13.50. 1 A 1500 nF capacitor is connected in parallel with a 16 Ωresistor across a 10 V, 10 kHz supply.

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Apr 28, 2015 · Because the 26 electrode layers are connected in parallel, the total resistance of N-layer shorting should equals 1/N of the shoring resistance for single-layer shorting, i.e. Figure 18 and Figure 19 display the calculated temperature rise of Al tab and SCO during the simulated internal short-circuit period.

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