In the network in Figure P2.11, the power absorbed by the resistor is 8 mW. Find VS. Figure P2.

In the network in Figure P2.11, the power absorbed by the...

Key Concepts

Voltage Source

A voltage source is a component that maintains a fixed potential difference across its terminals. It is critical in circuit theory because it provides the driving force for current flow, and its value can often be inferred from other known quantities in the circuit by applying circuit analysis techniques.

Circuit Analysis Techniques

These are systematic methods such as using Kirchhoff's Voltage and Current Laws or Thevenin's theorem to analyze complex circuits. They provide the necessary tools to relate various circuit parameters, facilitating the calculation of an unknown quantity, like the voltage of a source, by connecting known values such as the resistor's absorbed power.

Power Calculation

This concept refers to the process of determining the energy per unit time being dissipated, supplied, or absorbed by circuit elements using formulas such as P = V^2/R or P = I^2R. It is essential because power calculations allow for the determination of unknown voltages or currents in a network when the power is known.

Resistor Behavior

Resistors convert electrical energy into heat and their functioning is governed by Ohm's Law (V = IR). Understanding how the voltage across and the current through a resistor relate to its power absorption is key in analyzing and solving circuit problems involving resistors.

Transcript

00:01 Question number 99 according to the given circuit we will draw the diagram 5 oom resistance b to ampere c vc b resistance 10 oom i3crant a nought i1 5 volt source 1d resistance of 5 oom here the current so in this direction i2 point d 5 oham resistance i5 i4 resistance so 5 oom plus minus voltage source vs now this is the you can say uh uh circuit according to the given diagram so given power supplied is p supplied 2 is equal to 40 watt also from this figure v a d is equal to 5 volt v a d this is 5 volt hence p supplied 2 is equal to v c b into 2 v supplied here the value is 40 degree is 40 is equal to vc b multiply by 2 here we will get vc b is equal to 20 volts now let current in different branches i 1 i 2 i 2 i 2 i 3 i4 i5 are shown in this circuit diagram by oham's law oham's law v a b is equal to 2 into 5 which comes out v a b is equal to 10 volts by applying kirchof c .a.

05:05 V .a .b.

05:08 Plus v .c.

05:10 Plus v .c.

05:12 Is equal to 0 minus vca is equal to 10 minus 20.

05:21 Here, we will get vca is equal to minus 10 volt.

05:32 After that, by oam's law, vac, is equal to i3 into 10 vac value is minus 10 i 3 into 10 from here we will get the value of current 3 is equal to minus 1 ampere after that we will get the value of i 3 as minus 1 now by applying the kirchof rule, kirchof law at the loop acda.

06:47 So v -a -c plus v -c -d plus v -d -las -v -d -a is equal to 0.

06:57 As we have already calculated the value of v -a -c, which is minus 10, plus v -c -d, and the v -d -d -a -d -a is minus 5 is equal to 0.

07:12 From here, we will get vcd is equal to 15 volts...

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