Four lamps with unequal resistance values are connected in series to a 117-volt supply. If the t

Four lamps with unequal resistance values are connected in...

Key Concepts

Ohm's Law

Ohm's Law, expressed as V = IR, establishes the relationship between voltage, current, and resistance in an electrical circuit. It is used to determine how the total voltage is distributed across components and to calculate current when the total resistance is known.

Series Circuit

In a series circuit, all components are connected along a single path, meaning that the same current flows through every element regardless of their individual resistances. This principle is crucial to understanding why the current through each lamp is equal to the total current of the circuit.

Transcript

00:01 Hi, in the given problem there are four bulbs having different power ratings and which are arranged as shown here in this circuit diagram.

00:15 The first bulb, 60 watt power rating, the second in its parallel having 100 watt power rating then to another bulbs are joined.

00:31 In series with this parallel combination this one having 15 watt our rating and in front of it this one is having 40 watt power rating and combinedly the system is joined with the operating voltage of 120 volt so we have to find the net current coming out of this source i and we have to find the actual powers being consumed in these bulbs.

01:12 So the normal operating voltage for these bulbs if we consider it to be 120 volt and powers of these different bulbs are given to us here.

01:31 So, using the expression for the power as v squared by r, which gives us an expression for resistance of the bulb, r is equal to v square by p.

01:45 So first of all, we find the resistance of all the bulbs individually.

01:51 Here we consider this to be the first bulb, this we consider to be the second, this is the third one, and this is the fourth one.

02:02 So as per the order, r1 will be given by the normal operating voltage v squared divided by p1 means the square of 120 divided by 60 watt.

02:17 And this r1 here comes out to be 240 om.

02:23 In the similar manner for r2, for the second bulb, this is v square by p2 means 120 to the whole square by 100 which comes out to be 144 om then we find r3 for this 15 watt bulk that will be using the same formula 120 to the whole square divided by 15 and it will come out to be 960 om and finally for the fourth bulb r4 again 120 to the whole square divided by 40 watt and it will come out to be 360 oar so in order to find the total current coming out of the source first of all in the first part of the problem we reconstruct the circuit diagram showing the resistances replacing the bulbs with resistances here this will be r1 is equal to 240 om in its parallel this is r2 is equal to 144 om then the two bulbs which were in series in place of these bulbs we are showing their resistances r3 is equal to 960 om and finally this r4 is equal to 360 om and finally this r4 is equal to 360 om the source voltage 120 volt now here these r1 and r2 are in parallel first of all so their net resistance let it be r p will be given by 240 in 244 in the numerator their product and their addition in the denominator so this resistance the parallel combination comes out to be in 90 oom r p is equal to now this r3 r p and r4 all are in series so the net current coming out of this source i will be given by oms law v by net resistance which will be given by r3 plus r p plus r4 means this is 120 divided by 960 plus 90 plus 360 om.

05:17 So finally the answer for the first part of the problem here comes out to be 0 .085 ampere.

05:27 The current coming out of the source of this voltage...

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