Question

I need help with number 3 please!Parallel circuits, current, voltage, ohm's law

Answer 1

Given data:

* The voltage across the battery is,

`V_{\text{total}}=18\text{ V}`

Solution:

(a). Voltage:

The total voltage across the circuit is 18 V.

As the resistances are connected in parallel, thus, the voltage drop across each resistance is the same.

Hence, the voltage drop across the resistance 2 ohm is 18 V.

(b). Current:

The equivalent resistance of the circuit is,

`\begin{gathered} (1)/(R_(eq))=(1)/(R_1)+(1)/(R_2)+(1)/(R_3)+(1)/(R_4) \\ (1)/(R_(eq))=(1)/(2)+(1)/(3)+(1)/(6)+(1)/(9) \\ (1)/(R_(eq))=(9+6+3+2)/(18) \\ (1)/(R_(eq))=(20)/(18) \end{gathered}`

By simplifying,

`\begin{gathered} R_(eq)=(18)/(20) \\ R_(eq)=0.9\text{ ohm} \end{gathered}`

By Ohm's law, the total current through the circuit is,

`\begin{gathered} V_{\text{total}}=IR_(eq) \\ I=\frac{V_{\text{total}}}{R_(eq)} \\ I=(18)/(0.9) \\ I=20\text{ A} \end{gathered}`

Thus, the total current through the circuit is 20 A.

By Ohm's law, the current through the resistance 2 ohm is,

`\begin{gathered} V_{\text{total}}=I_1R_1 \\ I_1=\frac{V_{\text{total}}}{R_1} \\ I_1=(18)/(2) \\ I_1=9\text{ A} \end{gathered}`

Thus, the current through the resistance 2 ohm is 9 A.

(c). Resistance:

From part b, the total resistance of the circuit (equivalent resistance) is 0.9 ohm.

From the given values, the resistance R_1 is 2 ohm.

(d). Power:

The total power of the circuit is,

`\begin{gathered} P=V_{\text{total}}I \\ P=18*20 \\ P=360\text{ W} \end{gathered}`

Thus, the total power consumed by the circuit is 360 watts.

The power consumed by resistance R_1 is,

`\begin{gathered} P_1=I_1V_{\text{total}} \\ P_1=9*18 \\ P_1=162\text{ W} \end{gathered}`

Thus, the power consumed by resistance 2 ohm is 162 watts.