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Two uncharged capacitors with [tex]C_1 = 25 \mu F[/tex] and [tex]C_2 = 75 \mu F[/tex] are connected in parallel to a battery with [tex]\varepsilon = 7.5 \text{ V}[/tex].

Calculate the numerical value of [tex]Q_1[/tex] in [tex]\mu C[/tex].

Sagot :

GinnyAnswer
To find the charge [tex]\( Q_1 \)[/tex] on capacitor [tex]\( C_1 \)[/tex], we start with the given values for the capacitors and the battery voltage. The capacitance of [tex]\( C_1 \)[/tex] is [tex]\( 25 \: \mu F \)[/tex] and the voltage provided by the battery is [tex]\( 7.5 \: V \)[/tex].

We use the fundamental formula for the charge on a capacitor:
[tex]\[ Q = C \cdot V \][/tex]

Here, [tex]\( Q \)[/tex] is the charge on the capacitor, [tex]\( C \)[/tex] is the capacitance, and [tex]\( V \)[/tex] is the voltage across the capacitor.

Since the capacitors are connected in parallel, each capacitor experiences the full voltage provided by the battery. Therefore, the voltage [tex]\( V \)[/tex] across capacitor [tex]\( C_1 \)[/tex] is [tex]\( 7.5 \: V \)[/tex].

We substitute the given values into the formula:
[tex]\[ Q_1 = C_1 \cdot V \][/tex]
[tex]\[ Q_1 = 25 \: \mu F \cdot 7.5 \: V \][/tex]

Multiplying these together gives:
[tex]\[ Q_1 = 187.5 \: \mu C \][/tex]

So, the charge [tex]\( Q_1 \)[/tex] on capacitor [tex]\( C_1 \)[/tex] is [tex]\( 187.5 \: \mu C \)[/tex].
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