Current, I = 0.5 A   What is the cost of using the heater for 3 hours? What is the general name of the substances having infinitely high electrical resistance? (iii) Total resistance, RAC = RBC + RAC Now:

Three 2 V cells are connected in series and used as a battery in a circuit. So, the total resistance of the circuit can be calculated as: A current of 4 A flows around a circuit for 10 s. How much charge flows past a point in the circuit in this time? Thus, the resistance will be reduced by nine times.

Find the current in each resistor in the circuit shown below: In the given circuit, two 6 Ω and 4 Ω resistors are connected in parallel combination across a cell of 24 V. We know that in a parallel combination, the voltage across the resistor remains the same and only the current gets divided. Contains solved exercises, review questions, MCQs, important board questions and chapter overview. This free flow of electrons makes copper a good conductor of electricity. I(a) Let I1, I2 and I3 be the current flowing through the resistors of 5 Ω, 100 Ω and 30 Ω, respectively. (iii) the potential difference across R1. Here: (b) The current, I in the circuit can be calculated as: How many joules are there in one kilowatt-hour? Ohm's law gives the relationship between the current in a conductor and the potential difference across its ends.                                                                = 230 V × 0.4 A The substances that have infinitely high electrical resistance are called insulators. (b) In circuit (iii), the lamps will be as bright as in circuit (i) because circuit (iii) is a parallel arrangement. (d)40 Ω, (c) 20 Ω (a) 25 Ω  As shown  Resistance 2 Ω  and 3 Ω are in series we know resistance in series arrangement can be obtained as R=R1+R2                                                                                                 R=2 Ω+3 Ω=5 Ω Since this 5 Ω is in parallel with another 5 Ω resistance therefore total resistance can be obtained as1R=1R1+1R2   Here R1=5 Ω, R2=5 Ω1R=15+151R=25  R =2.5 ΩTotal resistance of combination=2.5 Ω A wire that has resistance R is cut into two equal pieces. The effective resistance of combination is 4 Ώ. (d) Draw a circuit diagram showing two electric lamps connected in parallel together with a cell and a switch that works both lamps. (a) the power of the kettle. 220 = 12 I (b) Length of the cable wire, l = 1 km = 1000 m. To find the resistance of the aluminium cable, R: Thus, the resistance of the copper wire, R = 86.5 Ω.                     R = 6/2 = 3 Ω 1/R = (1/5) + (1/10) + (1/30)

Energy consumed by the heater in 1 hour, E3 = 2 kWh 1/R' = (1/50) + (1/30) Thus, we will get a 1.5 Ω resistor.

         R2=3 Ω t = 1 s What is an ammeter?           R = 2203.82=57.60 Ω. (a) current and resistance For a heater rated at 4 kW and 220 V, calculate: The resistance, R' is connected in series with the two resistors of values 10 Ω each. (b) The resistance of a conductor is inversely proportional to its area of cross-section, i.e. Rate of electrical energy transfer = Power consumed The brightness of C will remain the same because the same current will continuously flow through bulb C. How do you think the brightness of two lamps arranged in parallel compares with the brightness of two lamps arranged in series (both arrangements having one cell)? x + y = 9                      ... (2) (f) A closed switch. The resistance of the light bulb is: (d) argon. In a series connection: (a) Draw a circuit diagram to represent this. (b) How much energy is transferred by a 12 V Power supply to each coulomb of charge which it moves around a circuit? (b) Parallel circuit

As we know from Ohm’s law: One watt can be defined as the power consumed by an electrical device if it is operated at a potential difference of 1 V and it carries a current of 1 A.

What is the SI unit of potential difference? If the bulb draws a current of 0.5 A, calculate the power of the bulb.

(c) volt A = area of the cross-section of the conductor (a) 30 A

It should be connected in series. We also share useful articles on our facebook page to help you in your board examination. I = V/R/2 = 2 I An electric heater of resistance 8 Ω takes a current of 15 A from the mains supply line.

Therefore, their net resistance can be calculated as: (iii) the total effective resistance of the circuit. Heat generated = I2Rt = (0.48)2(25)(60) = 345.6 J. Potential difference, V = 250/20 = 12.5 V Two gases that are filled in the filament type of electric bulbs are:

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