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Sagot :
To find the value of [tex]\( I_2 \)[/tex] given:
[tex]\[ R_1 = 9, \quad R_2 = 3, \quad I_1 = 15 \][/tex]
we will use the relationship based on Ohm's Law and the principles of parallel circuits.
1. Ohm's Law: The voltage drop across a resistor [tex]\( V \)[/tex] is given by
[tex]\[ V = I \cdot R \][/tex]
where [tex]\( I \)[/tex] is the current through the resistor and [tex]\( R \)[/tex] is the resistance of the resistor.
2. Parallel Circuits: In parallel circuits, the voltage across each branch is the same. Therefore, the voltage drop across [tex]\( R_1 \)[/tex] should be the same as the voltage drop across [tex]\( R_2 \)[/tex].
3. Using the current [tex]\( I_1 \)[/tex] through [tex]\( R_1 \)[/tex], calculate the voltage drop across [tex]\( R_1 \)[/tex]. This is given by:
[tex]\[ V = I_1 \cdot R_1 \][/tex]
4. Since the voltage drop across [tex]\( R_2 \)[/tex] is the same as the one across [tex]\( R_1 \)[/tex], we set this equal to the voltage drop across [tex]\( R_2 \)[/tex]. Let [tex]\( I_2 \)[/tex] be the current through [tex]\( R_2 \)[/tex]. Therefore,
[tex]\[ V = I_2 \cdot R_2 \][/tex]
5. We know that the voltage drops are equal, so:
[tex]\[ I_1 \cdot R_1 = I_2 \cdot R_2 \][/tex]
6. Substitute the values for [tex]\( R_1 \)[/tex], [tex]\( R_2 \)[/tex], and [tex]\( I_1 \)[/tex]:
[tex]\[ 15 \cdot 9 = I_2 \cdot 3 \][/tex]
7. Solve for [tex]\( I_2 \)[/tex]:
[tex]\[ 135 = I_2 \cdot 3 \][/tex]
Divide both sides by 3:
[tex]\[ I_2 = \frac{135}{3} \][/tex]
[tex]\[ I_2 = 45 \][/tex]
Therefore, the value of [tex]\( I_2 \)[/tex] is [tex]\( 45 \)[/tex].
[tex]\[ R_1 = 9, \quad R_2 = 3, \quad I_1 = 15 \][/tex]
we will use the relationship based on Ohm's Law and the principles of parallel circuits.
1. Ohm's Law: The voltage drop across a resistor [tex]\( V \)[/tex] is given by
[tex]\[ V = I \cdot R \][/tex]
where [tex]\( I \)[/tex] is the current through the resistor and [tex]\( R \)[/tex] is the resistance of the resistor.
2. Parallel Circuits: In parallel circuits, the voltage across each branch is the same. Therefore, the voltage drop across [tex]\( R_1 \)[/tex] should be the same as the voltage drop across [tex]\( R_2 \)[/tex].
3. Using the current [tex]\( I_1 \)[/tex] through [tex]\( R_1 \)[/tex], calculate the voltage drop across [tex]\( R_1 \)[/tex]. This is given by:
[tex]\[ V = I_1 \cdot R_1 \][/tex]
4. Since the voltage drop across [tex]\( R_2 \)[/tex] is the same as the one across [tex]\( R_1 \)[/tex], we set this equal to the voltage drop across [tex]\( R_2 \)[/tex]. Let [tex]\( I_2 \)[/tex] be the current through [tex]\( R_2 \)[/tex]. Therefore,
[tex]\[ V = I_2 \cdot R_2 \][/tex]
5. We know that the voltage drops are equal, so:
[tex]\[ I_1 \cdot R_1 = I_2 \cdot R_2 \][/tex]
6. Substitute the values for [tex]\( R_1 \)[/tex], [tex]\( R_2 \)[/tex], and [tex]\( I_1 \)[/tex]:
[tex]\[ 15 \cdot 9 = I_2 \cdot 3 \][/tex]
7. Solve for [tex]\( I_2 \)[/tex]:
[tex]\[ 135 = I_2 \cdot 3 \][/tex]
Divide both sides by 3:
[tex]\[ I_2 = \frac{135}{3} \][/tex]
[tex]\[ I_2 = 45 \][/tex]
Therefore, the value of [tex]\( I_2 \)[/tex] is [tex]\( 45 \)[/tex].
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