Question

Figure 27-78 shows a portion of a circuit through which there is a current I=6.00A. The resistances are R₁.=R₂=2.00R₃=2.OOR₄=4.00$\Omega$What is the currenti₁through resistor 1?

Short answer

The current i₁through resistor 1 is 2.0A.

Step-by-step solution

The given data

1. Current flowing through the circuit, I= 6.0A .
2. The relation between the resistances of the given resistors,$R_1=R_2=2R_3=2R_4=4.0\Omega$

Understanding the concept of current

If a circuit has more than one resistor, the current flowing through the circuit is given by first solving for the equivalence resistance of the circuit. Here, as we need the current through resistor 1, thus solving for the current for the parallel resistors 1 and 2 will provide the current value by symmetry relation between their resistances as their voltage will be equal.

Formulae:

The voltage equation using Ohm’s law,

V=IR (i)

The equivalent resistance for a series combination,

$R_{eq}=\underset{1}{\overset{n}{\sum R_i}}$ (ii)

The equivalent resistance for a parallel combination,

$R_{eq}=\underset{1}{\overset{n}{\sum \frac{1}{R_1}}}$ (iii)

Kirchhoff’s junction rule,

I_in = I_out (iv)

Here R is the resistance, and I is the current.

Calculation of the current through resistor 1

At first, the equivalence resistance of series resistors R₃and R₄can be given using equation (iii) as follows:

R₃₄ = R₃+R₄

Substitute the values in the above expression, and we get,

$$\begin{gathered} R_{34}=\frac{4.0\Omega }{2}+\frac{4.0\Omega }{2} \\ =4.0\Omega \end{gathered}$$

Now, the equivalence resistance of parallel resistors R₁and R₂ can be given using equation (ii) as follows:

$R_{12}=\left(R_1{R}_2\right)/(R_1+R_2)$

Substitute the values in the above expression, and we get,

localid="1662369087375" $$\begin{gathered} R_{12}=\frac{\left(4.0\Omega \left)\right(4.0\Omega \right)}{(4.0\Omega )+(4.0\Omega )} \\ =2.0\Omega \end{gathered}$$

From the figure, the voltage flowing through R₃₄and R₁₂must be equal for their parallel connection. Thus, the current equation can be given using equation (i) as follows:

$$\begin{gathered} V_{34}=V_{12} \\ {i}_{34}{R}_{34}=i_{12}{R}_{12} \\ {i}_{34}=\frac{i_{12}{R}_{12}}{R_{34}} \end{gathered}$$

Substitute the values in the above expression, and we get,

$\begin{array}{l}{i}_{34}=\frac{i_{12}\left(2.0\Omega \right)}{(4.0\Omega )}\\ =\frac{i_{12}}{2}\end{array}$

Now, using the junction rule of equation (iv), the current passing through resistors 1 and 2 can be given as follows:

I= i₁₂+i₃₄

Substitute the values in the above expression, and we get,

$\begin{array}{l}6.0A=i_{12}+\frac{i_{12}}{2}\\ \frac{3i_{12}}{2}=6.0A\\ i_{12}=4.0A\end{array}$

Now, as resistors 1 and 2 are in parallel connection and have equal resistance values; thus, the current flowing through resistor 1 can be given by symmetry as follows:

i₁=i_12/2

Substitute the values in the above expression, and we get,

$\begin{array}{l}{i}_1=\frac{4.0A}{2}\\ =2.0A\end{array}$

Hence, the value of current is 2.0A