From d49e31ba6756dbea2f79d7c538b5abd79a5f9054 Mon Sep 17 00:00:00 2001
From: Andrew Kesterson <andrew@aklabs.net>
Date: Thu, 4 Jun 2026 13:24:51 -0400
Subject: [PATCH] WIP

---
 05-photoresistors/README.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/05-photoresistors/README.md b/05-photoresistors/README.md
index 7b0bd90..410d99e 100644
--- a/05-photoresistors/README.md
+++ b/05-photoresistors/README.md
@@ -15,11 +15,11 @@ So when I looked at the circuit diagram, I wondered, "why are we measuring the v
 The answer is because, as it turns out, if you take the measurement downstream of the photoresistor in this particular example, you will always read 0v. You will never see a change in the output voltage from the photoresistor. In order to see the impact of the photoresistor in the circuit, you need to measure upstream of the photoresistor.
 
 <center>
-<img alt="wot" src="wot.jpeg"/>
-<br/>
-*The face of a man who realizes he has failed*
+<img alt="wot" width="320" src="wot.jpeg"/>
 
-*to grasp something very fundamental*
+<br/>
+<i>The face of a man who realizes he has failed<br/>
+to grasp something very fundamental</i>
 </center>
 
 If you remember back to [tutorial 4 with Analog Digital Conversion](../04-adc/README.md#measuring-potentiometers), I talked about potentiometers, and how they are a `voltage divider`.