For the capacitor network shown, the potential difference across ab is 63. a)find the total charge of the network b)find the charge on the 150nF capacitor c)find the charge on the 12nF capacitor d)FInd the total energy stored in the network e) Find the energy stored in the 150nf capacitor f)find the energy stored in the 120nF capacitor g)find the potential difference across the 150nf capacitor h) find the potential difference across the 120nf capacitor
For the capacitor network shown, the potential difference across ab is 63. a)find the total charge of the network b)find the charge on the 150nF capacitor c)find the charge on the 12nF capacitor d)FInd the total energy stored in the network e) Find the energy stored in the 150nf capacitor f)find the energy stored in the 120nF capacitor g)find the potential difference across the 150nf capacitor h) find the potential difference across the 120nf capacitor
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For the capacitor network shown, the potential difference across ab is 63.
a)find the total charge of the network
b)find the charge on the 150nF capacitor
c)find the charge on the 12nF capacitor
d)FInd the total energy stored in the network
e) Find the energy stored in the 150nf capacitor
f)find the energy stored in the 120nF capacitor
g)find the potential difference across the 150nf capacitor
h) find the potential difference across the 120nf capacitor
![The diagram illustrates two capacitors connected in series between points \( a \) and \( b \).
- The first capacitor, located closer to point \( a \), has a capacitance of 150 nanofarads (nF).
- The second capacitor, closer to point \( b \), has a capacitance of 120 nanofarads (nF).
The series connection implies that the total capacitance between points \( a \) and \( b \) can be calculated using the formula for capacitors in series:
\[
\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}
\]
where \( C_1 = 150 \) nF and \( C_2 = 120 \) nF. This arrangement is often used to understand the effect of combining different capacitors in an electrical circuit, which reduces the overall capacitance compared to the individual components.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe7d7e6ff-2bc1-4e75-b2e8-942a29e91675%2F1fe82c52-0c04-4c96-ab8e-90143c9a4fbd%2Fq4ifke9_processed.png&w=3840&q=75)
Transcribed Image Text:The diagram illustrates two capacitors connected in series between points \( a \) and \( b \).
- The first capacitor, located closer to point \( a \), has a capacitance of 150 nanofarads (nF).
- The second capacitor, closer to point \( b \), has a capacitance of 120 nanofarads (nF).
The series connection implies that the total capacitance between points \( a \) and \( b \) can be calculated using the formula for capacitors in series:
\[
\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}
\]
where \( C_1 = 150 \) nF and \( C_2 = 120 \) nF. This arrangement is often used to understand the effect of combining different capacitors in an electrical circuit, which reduces the overall capacitance compared to the individual components.
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