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Sagot :
To determine the relation between the currents in circuits [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we can proceed with the following steps:
1. Identify given values:
- Both circuits have identical voltage sources, so we can denote the voltage [tex]\( V \)[/tex] the same for both circuits.
- The resistor in circuit [tex]\( A \)[/tex] has a resistance [tex]\( R_A \)[/tex] which is half of the resistance [tex]\( R_B \)[/tex] in circuit [tex]\( B \)[/tex]. This gives us the equation:
[tex]\[ R_A = \frac{1}{2} R_B \][/tex]
2. Apply Ohm's Law:
- Ohm's Law states that [tex]\( V = I \cdot R \)[/tex]. We can solve this for current ([tex]\( I \)[/tex]):
[tex]\[ I = \frac{V}{R} \][/tex]
3. Calculate the currents in each circuit:
- For circuit [tex]\( A \)[/tex]:
[tex]\[ I_A = \frac{V}{R_A} \][/tex]
- For circuit [tex]\( B \)[/tex]:
[tex]\[ I_B = \frac{V}{R_B} \][/tex]
4. Express [tex]\( R_A \)[/tex] in terms of [tex]\( R_B \)[/tex]:
Since [tex]\( R_A = \frac{1}{2} R_B \)[/tex], substitute this into the equation for [tex]\( I_A \)[/tex]:
[tex]\[ I_A = \frac{V}{\frac{1}{2} R_B} = \frac{2V}{R_B} \][/tex]
5. Find the ratio of [tex]\( I_A \)[/tex] to [tex]\( I_B \)[/tex]:
- We already know that:
[tex]\[ I_B = \frac{V}{R_B} \][/tex]
- To compare [tex]\( I_A \)[/tex] and [tex]\( I_B \)[/tex], we take the ratio:
[tex]\[ \frac{I_A}{I_B} = \frac{\frac{2V}{R_B}}{\frac{V}{R_B}} = \frac{2V}{R_B} \times \frac{R_B}{V} = 2 \][/tex]
This shows that the current in circuit [tex]\( A \)[/tex] is 2 times the current in circuit [tex]\( B \)[/tex]. Therefore, the correct statement is:
C. The current flowing through circuit [tex]\( A \)[/tex] is 2 times as large as the current flowing through circuit [tex]\( B \)[/tex].
1. Identify given values:
- Both circuits have identical voltage sources, so we can denote the voltage [tex]\( V \)[/tex] the same for both circuits.
- The resistor in circuit [tex]\( A \)[/tex] has a resistance [tex]\( R_A \)[/tex] which is half of the resistance [tex]\( R_B \)[/tex] in circuit [tex]\( B \)[/tex]. This gives us the equation:
[tex]\[ R_A = \frac{1}{2} R_B \][/tex]
2. Apply Ohm's Law:
- Ohm's Law states that [tex]\( V = I \cdot R \)[/tex]. We can solve this for current ([tex]\( I \)[/tex]):
[tex]\[ I = \frac{V}{R} \][/tex]
3. Calculate the currents in each circuit:
- For circuit [tex]\( A \)[/tex]:
[tex]\[ I_A = \frac{V}{R_A} \][/tex]
- For circuit [tex]\( B \)[/tex]:
[tex]\[ I_B = \frac{V}{R_B} \][/tex]
4. Express [tex]\( R_A \)[/tex] in terms of [tex]\( R_B \)[/tex]:
Since [tex]\( R_A = \frac{1}{2} R_B \)[/tex], substitute this into the equation for [tex]\( I_A \)[/tex]:
[tex]\[ I_A = \frac{V}{\frac{1}{2} R_B} = \frac{2V}{R_B} \][/tex]
5. Find the ratio of [tex]\( I_A \)[/tex] to [tex]\( I_B \)[/tex]:
- We already know that:
[tex]\[ I_B = \frac{V}{R_B} \][/tex]
- To compare [tex]\( I_A \)[/tex] and [tex]\( I_B \)[/tex], we take the ratio:
[tex]\[ \frac{I_A}{I_B} = \frac{\frac{2V}{R_B}}{\frac{V}{R_B}} = \frac{2V}{R_B} \times \frac{R_B}{V} = 2 \][/tex]
This shows that the current in circuit [tex]\( A \)[/tex] is 2 times the current in circuit [tex]\( B \)[/tex]. Therefore, the correct statement is:
C. The current flowing through circuit [tex]\( A \)[/tex] is 2 times as large as the current flowing through circuit [tex]\( B \)[/tex].
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