10.0- resistor and a 40.0- resistor in series are connected to a 120-V dc source. (a) What is (are) the current(s) through the resistors current through the 10.0- resistor: A current through the 40.0- resistor: A (b) What is the voltage drop across each resistor voltage drop across the 10.0- resistor: V voltage drop across the 40.0-? resistor: V
The Correct Answer and Explanation is :
In a series circuit, the current flowing through each resistor is the same, while the voltage drop across each resistor depends on its resistance. Given a 10.0 Ω resistor and a 40.0 Ω resistor connected in series to a 120 V DC source, we can determine the current through the resistors and the voltage drop across each.
(a) Current through the resistors:
The total resistance ( R_{\text{total}} ) in a series circuit is the sum of the individual resistances:
[ R_{\text{total}} = R_1 + R_2 = 10.0\, \Omega + 40.0\, \Omega = 50.0\, \Omega ]
Using Ohm’s Law, ( V = I \times R ), we can solve for the current ( I ):
[ I = \frac{V}{R_{\text{total}}} = \frac{120\, \text{V}}{50.0\, \Omega} = 2.4\, \text{A} ]
Therefore, the current through both the 10.0 Ω and 40.0 Ω resistors is 2.4 A.
(b) Voltage drop across each resistor:
The voltage drop ( V ) across a resistor in a series circuit can be calculated using Ohm’s Law: ( V = I \times R ).
- Voltage drop across the 10.0 Ω resistor: [ V_{10} = I \times R_1 = 2.4\, \text{A} \times 10.0\, \Omega = 24.0\, \text{V} ]
- Voltage drop across the 40.0 Ω resistor: [ V_{40} = I \times R_2 = 2.4\, \text{A} \times 40.0\, \Omega = 96.0\, \text{V} ]
These voltage drops add up to the total supplied voltage:
[ V_{10} + V_{40} = 24.0\, \text{V} + 96.0\, \text{V} = 120.0\, \text{V} ]
This confirms that the calculations are consistent with the total voltage of the source.
In summary, when resistors are connected in series, the same current flows through each resistor, and the voltage drop across each resistor is proportional to its resistance. In this case, a 2.4 A current flows through both the 10.0 Ω and 40.0 Ω resistors, resulting in voltage drops of 24.0 V and 96.0 V, respectively.
