My science teacher posed a challenging question for us: “If half the resistance of a power grid is 128.3456713M ohms and the current is 30tc over 26 hours, calculate the potential difference and show all your workings.” I could really use some help with this! ðŸ˜
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english class
English Class Update We’re currently studying Romeo and Juliet in English, and honestly, it’s really confusing for me. The language feels like a jumbled mess, and my grades are suffering because of it. I really wish we could read a translated version so I could grasp the themes and symbolism better.
Certainly! Let’s tackle the problem step by step. We’ll start with the given values and use Ohm’s Law, which states that:
[ V = I \times R ]
where:
– ( V ) is the potential difference (voltage),
– ( I ) is the current (in amperes), and
– ( R ) is the resistance (in ohms).
Given:
Step 1: Calculate the Total Resistance
Since half of the resistance is given, we need to find the total resistance ( R ):
[
R = 2 \times 128.3456713 \text{ M}\Omega = 256.6913426 \text{ M}\Omega
]
Step 2: Convert the Total Resistance to Ohms
1 MΩ = ( 10^6 ) Ω, so:
[
R = 256.6913426 \times 10^6 \, \Omega = 256691342.6 \, \Omega
]
Step 3: Convert Current “tc” to Amperes
Without the exact value of “tc,” it’s a bit tricky to proceed. If you can provide the value of “tc” in terms of a common unit (like mA or A), that would help.
If we assume that tc is a typo or means “30 A:”, then we will use:
[
I = 30 \, \text{A}
]
Step 4: Calculate the Potential Difference
Now we can use Ohm’s Law to find the potential difference ( V ):
[
V = I \times R
]
Substituting the values we have:
[
V = 30 \, \text{A} \times 256691342.6 \, \Omega
]
Now we calculate ( V ):
[
V = 30 \times 256691342.6 = 7700740267.8 \, \text{V}
]
Step 5: Conclusion
So, the potential difference ( V ) (assuming a current of 30 A) across the power grid is approximately:
[
\boxed{7700740267.8 \, \text{V}}
]
(Note: Please clarify the unit for “tc” to ensure the accuracy of the current value. All steps are based on the assumption that tc might refer to amperes, but it’s important to confirm!)