[tex]\huge{ \mathfrak{ \underline{ Answer} \: \: ✓ }}[/tex]
1. Calculate the equivalent resistance in the circuit :
In the given diagram, the resistors are in series connection,
So equivalent resistance = sum of resistance of both resistors.
[tex] \large \boxed{R_{eq} =R_1 +R_2 }[/tex]
- [tex]R_{eq} = 480 + 360[/tex]
- [tex]R_{eq} = 840 \: ohms[/tex]
Therefore, Equivalent resistance = 840 ohms.
2. Calculate the current through the battery :
[tex] \boxed{ \mathrm{current = \dfrac{potential \: \: difference}{resistance} }}[/tex]
- [tex]i = \dfrac{120}{840} [/tex]
- [tex]i = \dfrac{1}{7} [/tex]
- [tex]i = 0.142 \: Amperes[/tex]
Hence, current through the battery = 0.142 A
3. How much current is passing through each resistor :
Since the resistors are joined in series connection, the current flowing through each resistor will be equal = 0.142 Amperes.
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[tex]\mathrm{ \#TeeNForeveR}[/tex]
