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Answer :
the resistance of the single resistor connected to the 9-V battery should be [tex]\( 7.5 \, \Omega \)[/tex].
To determine the resistance of a single resistor connected to a 9-V battery such that the current coming out of the battery matches the current in the circuit described in part d, we need to first calculate the current in the circuit in part d.
Given that the circuit in part d involves two 3-ohm resistors in parallel and then this combination in series with a 6-ohm resistor, we can calculate the equivalent resistance of this combination. The formula for calculating the equivalent resistance of resistors in parallel is:
[tex]\[ R_{\text{parallel}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} \][/tex]
In this case, [tex]\( R_1 = R_2 = 3 \, \Omega \)[/tex]. So,
[tex]\[ R_{\text{parallel}} = \frac{1}{\frac{1}{3} + \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 1.5 \, \Omega \][/tex]
Now, the equivalent resistance of the combination of the two 3-ohm resistors in parallel with the 6-ohm resistor in series is:
[tex]\[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = 1.5 + 6 = 7.5 \, \Omega \][/tex]
To calculate the current flowing through this circuit when connected to the 9-V battery, we can use Ohm's law:
[tex]\[ I = \frac{V}{R} \][/tex]
Where [tex]\( V = 9 \, \text{V} \)[/tex] and [tex]\( R = 7.5 \, \Omega \)[/tex].
[tex]\[ I = \frac{9}{7.5} = 1.2 \, \text{A} \][/tex]
So, the current flowing through the circuit in part d is [tex]\( 1.2 \, \text{A} \)[/tex].
Now, to find the resistance [tex](\( R_{\text{single}} \))[/tex] required for the single resistor connected to a 9-V battery to produce the same current, we use Ohm's law again:
[tex]\[ R_{\text{single}} = \frac{V}{I} \][/tex]
Where [tex]\( V = 9 \, \text{V} \)[/tex] and [tex]\( I = 1.2 \, \text{A} \)[/tex].
[tex]\[ R_{\text{single}} = \frac{9}{1.2} = 7.5 \, \Omega \][/tex]
So, the resistance of the single resistor connected to the 9-V battery should be [tex]\( 7.5 \, \Omega \)[/tex].
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