Answer :
The voltage drops across each resistor are as follows:
[tex](V_1 = 1\ \text{V})[/tex]
[tex](V_2 = 2\ \text{V})[/tex]
[tex](V_3 = 6\ \text{V})[/tex]
[tex](V_4 = 3\ \text{V})[/tex]
The total current flowing through the circuit is [tex](0.167\ \text{A}).[/tex]
The total resistance of the circuit is [tex](72\ \Omega).[/tex]
Total Resistance (RT):
The resistors are connected in series, so we can sum their resistances:[tex][ RT = R_1 + R_2 + R_3 + R_4 = 6\ \Omega + 12\ \Omega + 36\ \Omega + 18\ \Omega = 72\ \Omega ][/tex]
Total Current (IT):
We can use Ohm’s law to find the total current flowing through the circuit: [tex][ IT = \frac{V}{RT} = \frac{12\ \text{V}}{72\ \Omega} = 0.167\ \text{A} ][/tex]
Voltage Drops Across Each Resistor:
Now let’s calculate the voltage drops across each resistor:
[tex](V_1 = I_T \cdot R_1 = 0.167\ \text{A} \cdot 6\ \Omega = 1\ \text{V})[/tex]
[tex](V_2 = I_T \cdot R_2 = 0.167\ \text{A} \cdot 12\ \Omega = 2\ \text{V})[/tex]
[tex](V_3 = I_T \cdot R_3 = 0.167\ \text{A} \cdot 36\ \Omega = 6\ \text{V})[/tex]
[tex](V_4 = I_T \cdot R_4 = 0.167\ \text{A} \cdot 18\ \Omega = 3\ \text{V})[/tex]