Question

Rank in order, from largest to smallest, the currents $I_{\mathrm{a}}$to $I_{\mathrm{d}}$through the four resistors in FIGURE

Short answer

The currents are ranked from largest to smallest$I_c>I_a=I_d>I_b$.

Step-by-step solution

Step 1. Given information

Voltage is$V_a=2.0\Omega$

Resistance is $R_a=2.0\Omega$

Current $I_a=?$

Voltage $V_b=1.0\Omega$

Resistance $R_b=2.0\Omega$

Current ,$I_b=?$

Voltage $V_c=2.0\Omega$

Resistance $R_c=1.0\Omega$

Current$I_c=?$

Voltage $V_d=1.0\Omega$

Resistance $R_d=1.0\Omega$

Current $I_d=?$

Step 2. Find the currents are ranked from largest to smallest.

The following is the equation describing the relationship between current, applied voltage, and resistance:

$$\begin{gathered} V\propto I \\ V=IR \end{gathered}$$

The current is I, the applied voltage is V, and the resistance is R.

The following is the current equation derived from the above equation:

$I=\frac{V}{R}$

For resistor a,

The current flowing through resistor a is

$\begin{array}{rcl}{I}_a& =& \frac{2\mathrm{V}}{2\Omega }\\ & =& 1\mathrm{A}\end{array}$

For resistor b:

The current flowing through resistor b is

$\begin{array}{rcl}{I}_b& =& \frac{1\mathrm{V}}{2\Omega }\\ & =& 0.5\mathrm{A}\end{array}$

For resistor c:

The current flowing through resistor c is

$\begin{array}{rcl}{I}_c& =& \frac{2\mathrm{V}}{1\Omega }\\ & =& 2\mathrm{A}\end{array}$

For resistor d:

The current flowing through resistor d is

$\begin{array}{rcl}{I}_d& =& \frac{1\mathrm{V}}{1\Omega }\\ & =& 1\mathrm{A}\end{array}$

From the above calculation the order of current from largest to smallest is $I_c>I_a=I_d>I_b$