A DC circuit consists of a 20\text{ V} power source connected in series with a 4\text{ }\O… — Armed Services Vocational Aptitude Battery (ASVAB)

A DC circuit consists of a $20\text{ V}$ power source connected in series with a $4\text{ }\Omega$ resistor ( $R_1$ ). After passing through $R_1$ , the circuit splits into two parallel branches containing an $8\text{ }\Omega$ resistor ( $R_2$ ) and a $24\text{ }\Omega$ resistor ( $R_3$ ). The branches then recombine and connect back to the negative terminal of the power source. What is the electrical power dissipated by the $8\text{ }\Omega$ resistor ( $R_2$ )?

Cevap: 18 W

Cevap

18 Watts

To find the power dissipated by a single component in a compound circuit, you must isolate the parameters for that specific component. By finding the total resistance (

10\text{ }\Omega

) and total current (

2\text{ A}

), we can determine that the voltage dropping specifically across the parallel section is

12\text{ V}

. Because components in parallel share the same voltage, the

8\text{ }\Omega

resistor receives the full

12\text{ V}

. Using Watt's Law,

P = \frac{V^2}{R}

, we find

\frac{12^2}{8} = 18\text{ W}

Adım Adım Çözüm

Calculate the equivalent resistance of the parallel section (

R_p

R_p = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\text{ }\Omega

Before analyzing the whole circuit, parallel branches must be simplified into a single equivalent resistance.

Calculate the total resistance of the circuit (

R_t

R_t = 4\text{ }\Omega + 6\text{ }\Omega = 10\text{ }\Omega

The equivalent parallel resistance is in series with the

4\text{ }\Omega

resistor.

Calculate the total current (

I_t

) leaving the power source.

I_t = \frac{20\text{ V}}{10\text{ }\Omega} = 2\text{ A}

Ohm's law (

I = \frac{V}{R}

) relates the total source voltage to the total circuit resistance.

Calculate the voltage drop across the parallel section (

V_p

V_p = 2\text{ A} \times 6\text{ }\Omega = 12\text{ V}

The total current flows through the equivalent parallel resistance. This gives the voltage available specifically to the parallel branches.

Calculate the power dissipated by the

8\text{ }\Omega

resistor (

R_2

P = \frac{12^2}{8} = \frac{144}{8} = 18\text{ W}

Using Watt's Law (

P = \frac{V^2}{R}

), we apply the voltage specific to the parallel branch (

12\text{ V}

) and the individual resistance (

8\text{ }\Omega

Anahtar Kavram

Calculating power dissipation in a specific component of a series-parallel circuit requires finding the exact voltage or current acting on that specific component, not the entire circuit.

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru